Ask question

Ask question

All categories

4 years ago

11

Which value could be the length of the missing side of the triangle? (the side of the triangle has 5, the bottom has a 12 and th

e top has a X)
5
7
17
12

1 answer:

sveta [45]4 years ago

7 0

12 is the answer.

Hope it helps!

You might be interested in

Last question again someone pls help me I'm begging

Norma-Jean [14]

Answer:

Go for the 200 copies!

Step-by-step explanation:

Paying 77 times $0.39 is $30.03, which is more than the 200 copies offer, so even if you throw away all the copies you don't need, the 200 offer is still cheaper!

4 0

2 years ago

Read 2 more answers

Can someone answer 1 and 2 for me? thanks so much in advance :p

mrs_skeptik [129]

i can help you with 1 .Sorry that i cant do 2 because number 2 doesnt have all the question in the photo.

1 A:No
1 B:Yes
1 C:Yes

4 0

4 years ago

Read 2 more answers

A 7-foot board is cut into two pieces. If one piece is x feet long, express the other length in terms of X

goldfiish [28.3K]

Answer:

y = 7 - x

Step-by-step explanation:

Let y be the length of the other section

x + y = 7

y = 7 - x

of course, the width of the saw kerf (k) should also be accounted for so a carpenter would figure it as

y = 7 - x - k

6 0

3 years ago

The two bases of a trapezoid are 3 and 11 and the altitude is 8 . What is the area of the trapezoid?

cluponka [151]

We are given two bases as 3 and 11

altitude is 8

formula for area of trapezoid is given by

$A=\frac{a+b}{2} *h$

here a and b are two bases and h is height

plugging the given values we get

$A=\frac{11+3}{2} *8$ =56

Area of trapezoid = 56 unit square

5 0

3 years ago

A rectangular field with perimeter of 80m is to have an area of at least 380m² . Describe the possible lengths and of the field.

Kryger [21]

$let \: x \: be \: the \: length \\ let \: y \: be \: the \: width$

$perimeter = 2x + 2y \\ area = xy$

$2x + 2y = 80 \\ xy = 380 \\ \\ x + y = 40 \\ xy = 380 \\ \\ x = 40 - y \\ (40 - y)y = 380$

$40y - y {}^{2} = 380 \\ y {}^{2} - 40y + 380 = 0 \\ y {}^{2} - 40y + 400 = 20 \\ (y - 20) {}^{2} = 20 \\ y - 20 =± \sqrt{20} \\ y_{1}=20-√20≈15.52786\\ y_{2}=20+√20≈24.4721$

6 0

2 years ago