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3 years ago

Need help ASAP Please

Mathematics

2 answers:

Eduardwww [97]3 years ago

5 0

Answer:

area: $16x^2+24x+9$ Perimeter: 16x+12

Step-by-step explanation:

area : $(4x+3)^2$

thats $16x^2+24x+9$

P: 4(4x+3)

thats 16x+12

Hope this helped you

Red

White raven [17]3 years ago

3 0

Answer:

Perimeter is all sides added together, while Area is one side height times length

Step-by-step explanation:

Since your shape is a square all sides are equal so the equation is the same for all sides

Perimeter = 4z+3+4z+3+4z+3+4z+3 or (4z+3)4

14z+12 <= answer for perimeter

Area = (4z+3) (4z+3)

16z^2+12z+12z+9

16z^2+24z+9 <= answer for area

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3 years ago

Read 2 more answers

The Venn diagram shows the results of two events resulting from rolling a number cube.

tankabanditka [31]

Answer:

$P(A|B)=\frac{2}{3}$

$P(A)*P(B)=\frac{1}{3}$

$P(A) =\frac{2}{3}$

$P(B) =\frac{1}{2}$ .

Step-by-step explanation:

We use the Venn diagram to calculate the desired probabilities.

Note that there are 6 possible results in the sample space

S = {1, 2, 3, 4, 5, 6}

Then note that in the region representing the intercept of A and B there are two possible values.

$P (A\ and\ B) = \frac{2}{6} = \frac{1}{3}$

In the region that represents event A there are 4 possible outcomes {4, 5, 1, 2}

$P(A) = \frac{4}{6} = \frac{2}{3}$

In the region that represents event B there are 3 possible outcomes {1, 2, 6}

$P(B) = \frac{3}{6} = \frac{1}{2}$ .

Now

$P(A | B)=\frac{P(A \ and\ B)}{P(B)}\\\\P(A | B)=\frac{\frac{1}{3}}{\frac{1}{2}}\\\\P(A|B)=\frac{2}{3}$

$P(A)*P(B)=\frac{2}{3}*\frac{1}{2}=\frac{1}{3}$

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3 years ago

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Lana71 [14]

We have to find how can someone convert the temperature from degrees Farhenheit to degrees Celsius using the function. The temperature in degrees Celsius is equal to the temperature in degrees Farhenheit minus 32 times 5/9. Or C ( F ) = ( F - 32 ) * 5/9. For example: the temperature is F = 50°; C ( 50 ) = ( 50 - 32 ) * 5/9 = 18 * 5/ 9 = 10 °C. Answer: The function is: C ( F ) = ( F - 32 ) * 5/9.

Hope this helped.

6 0

3 years ago

Read 2 more answers

I NEED HELP PLEASEE

xenn [34]

$\bf \cfrac{1+cot^2(\theta )}{1+csc(\theta )}=\cfrac{1}{sin(\theta )} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{1+cot^2(\theta )}{1+csc(\theta )}\implies \cfrac{1+\frac{cos^2(\theta )}{sin^2(\theta )}}{1+\frac{1}{sin(\theta )}}\implies \cfrac{~~\frac{sin^2(\theta )+cos^2(\theta )}{sin^2(\theta )}~~}{\frac{sin(\theta )+1}{sin(\theta )}}\implies \cfrac{~~\frac{1}{sin^2(\theta )}~~}{\frac{sin(\theta )+1}{sin(\theta )}}$

$\bf \cfrac{1}{\underset{sin(\theta )}{~~\begin{matrix} sin^2(\theta ) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }}\cdot \cfrac{~~\begin{matrix} sin(\theta ) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{sin(\theta )+1}\implies \cfrac{1}{sin^2(\theta )+sin(\theta )} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{1+cot^2(\theta )}{1+csc(\theta )}\ne \cfrac{1}{sin(\theta )}~\hfill$

5 0

3 years ago