I need you guy’s help answer thanks so much

We know that [tex]\log_a(bc)=\log_a(b)+\log_a(c)[/tex].

Using this rule,

[tex]\log_c(A^5B^3)=\log_c(A^5)+\log_c(B^3)[/tex].

We also know that [tex]\log_c(a^b)=b\log_c(a)[/tex].

Using this rule,

[tex]\log_c(A^5)+\log_c(B^3)=5\log_c(A)+3\log_c(B)[/tex]

Now we know that [tex]\log_c(A)=2,\log_c(B)=5[/tex] so,

[tex]5\cdot2+3\cdot5=10+15=\boxed{25}[/tex].

Hope this helps :)

Excel enables the users to perform mathematics basic and advanced function with just one formula.

The formula for sum of entire row or column can be done with just entering a single formula and results are shown in seconds.

The formula for sum of few column cells is,

=SUM(B2:B6)

The spreadsheet allows the user to enter various formula and results are displayed withing seconds.

There are formulas for basic math functions and there are also formulas for advance mathematics calculations. For addition of values of many cells sum formula is used and range is assigned for reference.

The formula adds all the values of selected cells and displays the results in different cell.

Learn more at https://brainly.com/question/24365931

Answer:

[tex]{ \sf{ {x}^{2} + {y}^{2} = {(x + y)}^{2} - 2xy }}[/tex]

Step-by-step explanation:

[tex]{ \tt{ {(x + y)}^{2} = (x + y)(x + y) }} \\ { \tt{ {(x + y)}^{2} = ( {x}^{2} + 2xy + {y}^{2}) }} \\ { \tt{( {x}^{2} + {y}^{2} ) = {(x + y)}^{2} - 2xy}}[/tex]

Answer: 17 feet

Step-by-step explanation:

51/48 = x/16

(51)(16)/48

The statute is 17 feet tall.

Similar triangles are the triangles that have the same shape, but different sizes. The corresponding angles are congruent and the sides are in proportion.

The ratio of any corresponding sides in two equiangular triangles is always the same.

Let's visualize the situation according to the given question.

AB is the building ,whose height is 51f

BC is the shadow of the building AB, whose length is 48ft.

QR is the shadow of the tower statue, whose length is 16feet.

Let the height of the statue PR be h feet.

In triangle ACB and triangle PRQ

∠ACB = ∠PRQ = 90 degrees  

( the objects and shadows are perpendicular to each other)

∠BAC = ∠QPR

( sunray falls on the pole and tower at the same angle, at the same time )

⇒ΔACB similar to ΔPRQ   ( AA criterion)

Therefore, the ratio of any two corresponding sides in equiangular triangles is always same.

⇒ AC/CB = PR/RQ

⇒[tex]\frac{51}{48} =\frac{h}{16}[/tex]

⇒ h = [tex]\frac{(51)(16)}{48}[/tex]

⇒ h = 17 feet.

Hence, the statute is 17 feet tall.    

Learn more about the similar triangle here:

brainly.com/question/25882965

#SPJ2

Step-by-step explanation:

\underline{\textsf{Given:}}

Given:

\mathsf{Polynomial\;is\;x^3+7x^2+7x-15}Polynomialisx

3

+7x

2

+7x−15

\underline{\textsf{To find:}}

To find:

\mathsf{Factors\;of\;x^3+7x^2+7x-15}Factorsofx

3

+7x

2

+7x−15

\underline{\textsf{Solution:}}

Solution:

\textsf{Factor theorem:}Factor theorem:

\boxed{\mathsf{(x-a)\;is\;a\;factor\;P(x)\;\iff\;P(a)=0}}

(x−a)isafactorP(x)⟺P(a)=0

\mathsf{Let\;P(x)=x^3+7x^2+7x-15}LetP(x)=x

3

+7x

2

+7x−15

\mathsf{Sum\;of\;the\;coefficients=1+7+7-15=0}Sumofthecoefficients=1+7+7−15=0

\therefore\mathsf{(x-1)\;is\;a\;factor\;of\;P(x)}∴(x−1)isafactorofP(x)

\mathsf{When\;x=-3}Whenx=−3

\mathsf{P(-3)=(-3)^3+7(-3)^2+7(-3)-15}P(−3)=(−3)

3

+7(−3)

2

+7(−3)−15

\mathsf{P(-3)=-27+63-21-15}P(−3)=−27+63−21−15

\mathsf{P(-3)=63-63}P(−3)=63−63

\mathsf{P(-3)=0}P(−3)=0

\therefore\mathsf{(x+3)\;is\;a\;factor}∴(x+3)isafactor

\mathsf{When\;x=-5}Whenx=−5

\mathsf{P(-5)=(-5)^3+7(-5)^2+7(-5)-15}P(−5)=(−5)

3

+7(−5)

2

+7(−5)−15

\mathsf{P(-5)=-125+175-35-15}P(−5)=−125+175−35−15

\mathsf{P(-5)=175-175}P(−5)=175−175

\mathsf{P(-5)=0}P(−5)=0

\therefore\mathsf{(x+5)\;is\;a\;factor}∴(x+5)isafactor

\underline{\textsf{Answer:}}

Answer:

\mathsf{x^3+7x^2+7x-15=(x-1)(x+3)(x+5)}x

3

+7x

2

+7x−15=(x−1)(x+3)(x+5)

\underline{\textsf{Find more:}}

Find more:

Answer:

r + s - 19

General Formulas and Concepts:

Algebra I

Step-by-step explanation:

Step 1: Define

Identify

r - 7 + s - 12

Step 2: Simplify

Answer:

18c^3d^9

Step-by-step explanation:

2c^3 d^2 * 9d^7

We know that  we add the exponents when the base is the same

2*9  c^3 d^(2+7)

18c^3d^9

Answer:

if the zeros are x = -2, x = 0, and x = 1

then (x + 2), x and (x - 1) are factors of the polynomial

multiply these factors together

p(x) = x(x + 2)(x - 1)

p(x) = x(x2 + x - 2)

p(x) = x3 + x2 - 2x

this is a polynomial of degree 3 with the given roots

Answer:

Center: (1,3)

Radius: 6

Step-by-step explanation:

Hi there!

[tex]x^2-2x + y^2 - 6y = 26[/tex]

Typically, the equation of a circle would be in the form [tex](x-h)^2+(y-k)^2=r^2[/tex] where [tex](h,k)[/tex] is the center and [tex]r[/tex] is the radius.

To get the given equation [tex]x^2-2x + y^2 - 6y = 26[/tex] into this form, we must complete the square for both x and y.

1) Complete the square for x

Let's take a look at this part of the equation:

[tex]x^2-2x[/tex]

To complete the square, we must add to the expression the square of half of 2. That would be 1² = 1:

[tex]x^2-2x+1[/tex]

Great! Now, let's add this to our original equation:

[tex]x^2-2x+1+y^2-6y = 26[/tex]

We cannot randomly add a 1 to just one side, so we must do the same to the right side of the equation:

[tex]x^2-2x+1+y^2-6y = 26+1\\x^2-2x+1+y^2-6y = 27[/tex]

Complete the square:

[tex](x-1)^2+y^2-6y = 27[/tex]

2) Complete the square for y

Let's take a look at this part of the equation [tex](x-1)^2+y^2-6y = 27[/tex]:

[tex]y^2-6y[/tex]

To complete the square, we must add to the expression the square of half of 6. That would be 3² = 9:

[tex]y^2-6y+9[/tex]

Great! Now, back to our original equation:

[tex](x-1)^2+y^2-6y+9= 27[/tex]

Remember to add 9 on the other side as well:

[tex](x-1)^2+y^2-6y+9= 27+9\\(x-1)^2+y^2-6y+9= 36[/tex]

Complete the square:

[tex](x-1)^2+(y-3)^2= 36[/tex]

3) Determine the center and the radius

[tex](x-1)^2+(y-3)^2= 36[/tex]

[tex](x-h)^2+(y-k)^2=r^2[/tex]

Now, we can see that (1,3) is in the place of (h,k). 36 is also in the place of r², making 6 the radius.

I hope this helps!

Answer:

[tex]\sqrt{g^2+f^2-c}[/tex]

[tex]g=-1,f=-3,c=-26[/tex]

so, the Center of the equation is [tex](1,3)[/tex]

[tex]\sqrt{(-1)^2+(-3)^2-(-26})[/tex]

[tex]=\sqrt{1+9+26}[/tex]

[tex]=\sqrt{36}[/tex]

[tex]=6[/tex]

OAmalOHopeO

Answer:

C

Step-by-step explanation:

Answer:

Hence the adjusted R-squared value for this model is 0.7205.

Step-by-step explanation:

Given n= sample size=20  

Total Sum of square (SST) =1000  

Model sum of square(SSR) =750  

Residual Sum of Square (SSE)=250  

The value of R ^2 for this model is,  

R^2 = \frac{SSR}{SST}  

R^2 = 750/1000 =0.75  

Adjusted [tex]R^2[/tex] :

[tex]Adjusted R^2 =1- \frac{(1-R^2)\times(n-1)}{(n-k-1)}[/tex]

Where k= number of regressors in the model.

[tex]Adjusted R^2 =1-(19\times 0.25/((20-2-1)) = 0.7205[/tex]

Answer:

See pic below.

Answer: C. 380°

Step-by-step explanation:

Answer:

Mean of sampling distribution = 50

Standard deviation of sampling distribution, = 0.625

Step-by-step explanation:

Given :

Mean, μ = 50

Standard deviation, σ = 5

Sample size, n = 64

The mean of sampling distribution, μxbar = population mean, μ

μxbar = μ

According to the central limit theorem, the sampling distribution converges to the population mean as the sample size increases, hence, , μxbar = μ = 50

Standard deviation of sampling distribution, σxbar = σ/√n

σxbar = 5/√64 = 5 / 8 = 0.625

9514 1404 393

Answer:

  186

Step-by-step explanation:

Let b represent the number of boxes in the truck. Then for the weight limit to be met, we require ...

  3512 +16b < 6500

  16b < 2988

  b < 186.75

The maximum number of boxes is 186.

Answer:

Step-by-step explanation:

Replace (x, y) with (1, 1) and verify if the equations are correct.

You would ignore the x and y in the equations. Testing one of each pair.

It is obvious that only b. is correct.

Answer:

Step-by-step explanation:

Here's how you convert:

[tex]\sqrt[n]{x^m}=x^{\frac{m}{n}[/tex]  The little number outside the radical, called the index, serves as the denominator in the rational power, and the power on the x inside the radical serves as the numerator in the rational power on the x.

A couple of examples:

[tex]\sqrt[3]{x^4}=x^{\frac{4}{3}[/tex]

[tex]\sqrt[5]{x^7}=x^{\frac{7}{5}[/tex]

It's that simple. For your problem in particular:

[tex]\sqrt[3]{7}[/tex] is the exact same thing as [tex]\sqrt[3]{7^1}=7^{\frac{1}{3}[/tex]

Solution :

Let [tex]p_1[/tex] and [tex]p_2[/tex]  represents the proportions of the seeds which germinate among the seeds planted in the soil containing [tex]3\%[/tex] and [tex]5\%[/tex] mushroom compost by weight respectively.

To test the null hypothesis [tex]H_0: p_1=p_2[/tex] against the alternate hypothesis  [tex]H_1:p_1 \neq p_2[/tex] .

Let [tex]\hat p_1, \hat p_2[/tex] denotes the respective sample proportions and the [tex]n_1, n_2[/tex] represents the sample size respectively.

[tex]$\hat p_1 = \frac{74}{155} = 0.477419[/tex]

[tex]n_1=155[/tex]

[tex]$p_2=\frac{86}{155}=0.554839[/tex]

[tex]n_2=155[/tex]

The test statistic can be written as :

[tex]$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}[/tex]

which under [tex]H_0[/tex]  follows the standard normal distribution.

We reject [tex]H_0[/tex] at [tex]0.05[/tex] level of significance, if the P-value [tex]<0.05[/tex] or if [tex]|z_{obs}|>Z_{0.025}[/tex]

Now, the value of the test statistics = -1.368928

The critical value = [tex]\pm 1.959964[/tex]

P-value = [tex]$P(|z|> z_{obs})= 2 \times P(z< -1.367928)$[/tex]

                                     [tex]$=2 \times 0.085667$[/tex]

                                     = 0.171335

Since the p-value > 0.05 and [tex]$|z_{obs}| \ngtr z_{critical} = 1.959964$[/tex], so we fail to reject [tex]H_0[/tex] at [tex]0.05[/tex] level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the [tex]\text{proportion}[/tex] of the seeds that [tex]\text{germinate differs}[/tex] with the percent of the [tex]\text{mushroom compost}[/tex] in the soil.

9514 1404 393

Answer:

  a) 16A -4B +C = 1

  b) 4A -2B +C = 0.94

  c) C = 1

Step-by-step explanation:

Substitute the x- and y-values into the general form equation.

a. A(-4)² +B(-4) +C = 1

  16A -4B +C = 1

__

b. A(-2)² +B(-2) +C = 0.94

  4A -2B +C = 0.94

__

c. A(0)² +B(0) +C = 1

  C = 1

_____

Additional comment

Solving these equations gives A=0.015, B=0.06, C=1. The quadratic is ...

  0.015x² +0.06x +1 = y

Area of rhombus = 1/2 × d1 × d2

Let the other diagonal be x

ATQ

1/2 × 18 × x = 24

9 × x = 24

x = 24/9

x = 8/3

Now half each diagonal = 9 and 4/3

Now side = b² + p² = h²

9²+(4/3)² = h²

81 + 16/9 = h²

729/9 + 16/9 = h²

745/9 = h²

√(745/9) = h

Therefore the side of the rhombus is √(745/9)cm

Answered by Gauthmath must click thanks and mark brainliest

9514 1404 393

Answer:

  (a) The blue line ... solution ... (-6, -2)..

Step-by-step explanation:

The second equation describes a line with negative slope and a y-intercept of -4. This is clearly the red line on the graph.

The blue line represents the equation 2x -3y = -6.

The point of intersection of the two lines is (-6, -2), so that is the solution to the system of equations. This, by itself, is sufficient for you to choose the correct answer.

Answer:

0.6 = 60% probability that a randomly selected application from this distribution took less than 18 minutes.

Step-by-step explanation:

Uniform probability distribution:

An uniform distribution has two bounds, a and b.  

The probability of finding a value of at lower than x is:

[tex]P(X < x) = \frac{x - a}{b - a}[/tex]

The probability of finding a value between c and d is:

[tex]P(c \leq X \leq d) = \frac{d - c}{b - a}[/tex]

The probability of finding a value above x is:

[tex]P(X > x) = \frac{b - x}{b - a}[/tex]

Uniform distribution between 15 and 20 minutes.

This means that [tex]a = 15, b = 20[/tex]

What is the probability that a randomly selected application from this distribution took less than 18 minutes?

[tex]P(X < 18) = \frac{18 - 15}{20 - 15} = 0.6[/tex]

0.6 = 60% probability that a randomly selected application from this distribution took less than 18 minutes.

Answer:

Feet to Miles

Step-by-step explanation:

Answer:

Feet to Miles

Step-by-step explanation:

Both units must measure the same quantity.

Answer:

The point estimate for the proportion of cell phone owners aged 18-24 who have an Android phone is 0.4137.

Step-by-step explanation:

The point estimate is the sample proportion.

Sample proportion:

103 out of 249, so:

[tex]p = \frac{103}{249} = 0.4137[/tex]

The point estimate for the proportion of cell phone owners aged 18-24 who have an Android phone is 0.4137.

Answer:

You'd think its 34% but apparently it's 16%.

I hope this is right. If its not then it must be 34%.

(A) -- or maybe (B). 80% confident it is A.

ED2021

Answer:

"D"

if you multiply by Conjugate

the denominator would end up A^2 - b^2

the answer has 25 - 10x

that is D

Step-by-step explanation:

(D)

Step-by-step explanation:

Multiply and divide the fraction by the conjugate:

[tex]\dfrac{2}{\sqrt{13}+\sqrt{11}}×\dfrac{\sqrt{13}-\sqrt{11}}{\sqrt{13}-\sqrt{11}}[/tex]

[tex]= \dfrac{2(\sqrt{13} - \sqrt{11})}{(\sqrt{13})^2 - (\sqrt{11})^2}[/tex]

[tex]=\dfrac{2(\sqrt{13} - \sqrt{11})}{2}[/tex]

[tex]=\sqrt{13} - \sqrt{11}[/tex]

Answer:

Step-by-step explanation: