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2 years ago

13

Find the surface area of the rectangular prism.

1 answer:

AleksAgata [21]2 years ago

6 0

The surface area is

2(5*4)+2(5*2)+2(4*2)=

40+20+16=

76square yards

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Find the distance between points M(6,16) and Z(-1,14) to the nearest tenth.

Brut [27]

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
M&({{ 6}}\quad ,&{{ 16}})\quad 
% (c,d)
Z&({{ -1}}\quad ,&{{ 14}})
\end{array}\quad 
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
MZ=\sqrt{(-1-6)^2+(14-16)^2}\implies MZ=\sqrt{(-7)^2+(-2)^2}
\\\\\\
MZ=\sqrt{49+4}\implies MZ=\sqrt{53}\implies MZ\approx 7.3$

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4 years ago

Jack bought a new coat that was $120 but he had a 35% off coupon. If he had to pay 8% for tax, what

Elena-2011 [213]

Answer:

80% im not sure if you wanted that or the number of the price‍♀️

5 0

3 years ago

Read 2 more answers

I need help mainly the anwsers

Angelina_Jolie [31]

Answer:

3(-2)+2= -4

3(3)+2= 11

3(-0.5)+2= .5

3(1 1/2)+2=6.5

Step-by-step explanation:

just plug them in as x and you get the answer

3 0

3 years ago

Solve the formula <br> d=rt for t<br> .

vredina [299]

Answer:

D/R=T

Step-by-step explanation:

Divide both sides by r

4 0

3 years ago

Read 2 more answers

The sum of the squares of two numbers is 8 . The product of the two numbers is 4 . Find the numbers.

fenix001 [56]

Hello there.

First, assume the numbers $x,~y$ such that they satisties both affirmations:

The sum of the squares of two numbers is $8$ .
The product of the two numbers is $4$ .

With these informations, we can set the following equations:

$\begin{center}\align x^2+y^2=8\\ x\cdot y=4\\\end{center}$

Multiply both sides of the second equation by a factor of $2$ :

$2\cdot x\cdot y = 2\cdot 4\\\\\\ 2xy=8~~~~~(2)^{\ast}$

Make $(1)-(2)^{\ast}$

$x^2+y^2-2xy=8-8\\\\\\ x^2-2xy+y^2=0$

We can rewrite the expression on the left hand side using the binomial expansion in reverse: $(a-b)^2=a^2-2ab+b^2$ , such that:

$(x-y)^2=0$

The square of a number is equal to $0$ if and only if such number is equal to $0$ , thus:

$x-y=0\\\\\\ x=y~~~~~~(3)$

Substituting that information from $(3)$ in $(2)$ , we get:

$x\cdot x = 4\\\\\\ x^2=4$

Calculate the square root on both sides of the equation:

$\sqrt{x^2}=\sqrt{4}\\\\\\ |x|=2\\\\\\ x=\pm~2$

Once again with the information in $(3)$ , we have that:

$y=\pm~2$

The set of solutions of that satisfies both affirmations is:

$S=\{(x,~y)\in\mathbb{R}^2~|~(x,~y)=(-2,\,-2),~(2,~2)\}$

This is the set we were looking for.

7 0

3 years ago