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3 years ago

Help me please this is donr tomorrow and if i don't finish it my teacher will be so mad

Mathematics

1 answer:

Genrish500 [490]3 years ago

4 0

3/4 since (9 1/2 = length* width with makes the base) and divide it by 7 1/8.<span />

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Please help me! I am very confused

Sphinxa [80]

Answer:

1) 22.5 A

2) 112 B

3) 430 L

4) 576 P

5) 486 L

6) 624 S

Code is A B L P L S

Step-by-step explanation:

1) Area of a triangle = 1/2 * base * height = 1/2 *5*9 = 22.5

2) Area of a parallelogram = base x height = 14 x 8 = 112

3) Area of a rectangular prism = 2(length * width) + 2*(length + height) + 2 *(length *height)

= 2(15x7 + 15*5 + 7*5)

= 2(105 + 75 + 35)

= 2 * 215

= 430

4) Volume of a triangular prism = 1/2 * base * length * height

= 1/2 * 8 * 16 * 9

= 576

5) A cube has six surfaces. Each surface has an area of s x s where s is the length of each side. In this case, each side has area of 9x9 = 81. Total surface area = 6 x 81 = 486 and that is the paper required

6) The trailer is a rectangular prism so its volume = length x width x height = 13 6 x 8 = 624

Now you have to look at each value and see which letter it corresponds to. For example answer 1) is 22.5 which lies between 0-100 so it gets letter A, answer (2) is 112 which lies in the range 101-200 so it gets the letter B and so on

4 0

2 years ago

Can someone please help me ASAP

butalik [34]

Answer:

option D is the correct answer.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

How many sides does 52 quadrilateral have?

pickupchik [31]

52 quadrilaterals would have 52 ÷ 4 = a
a = your answer.
My answer is reasonable because if you have 52 sides from quadrilaterals you would need to divide by 4 to get the amount of quadrilaterals you have. Check your work by multiplying 4 x a = __ (The blank should be 52)

5 0

3 years ago

There are 12,boys and 10 girls in your gym class. If 6 boys joined the class, how many girls would need to join for the ratio to

Savatey [412]

6 girls would need to join for the ratio to be the same

5 0

3 years ago

Read 2 more answers

.. Which of the following are the coordinates of the vertices of the following square with sides of length a?

atroni [7]

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Step-by-step explanation:

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

To find the sides of a square, let us use the distance formula,

$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } S T=\sqrt{(a-0)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } T W=\sqrt{(a-a)^{2}+(0-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,a), T(a,a), W(a,0)$ has sides of length a.

Option B: $O(0,0), S(0,a), T(2a,2a), W(a,0)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text {Length } S T=\sqrt{(2 a-0)^{2}+(2 a-a)^{2}}=\sqrt{5 a^{2}}=a \sqrt{5}\\&\text {Length } T W=\sqrt{(a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{2 a^{2}}=a \sqrt{2}\\&\text {Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

This is not a square because the lengths are not equal.

Option C: $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length OS }=\sqrt{(0-0)^{2}+(2 a-0)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } S T=\sqrt{(2 a-0)^{2}+(2 a-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } T W=\sqrt{(2 a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } O W=\sqrt{(2 a-0)^{2}+(0-0)^{2}}=\sqrt{4 a^{2}}=2 a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$ has sides of length 2a.

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length OS }=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } S T=\sqrt{(a-a)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } T W=\sqrt{(0-a)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } O W=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

Thus, the square with vertices $O(0,0), S(a,0), T(a,a), W(0,a)$ has sides of length a.

Thus, the correct answers are option a and option d.

8 0

3 years ago