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3241004551 [841]
3 years ago
13

LAST QUESTION PLEASE HELP!! Which expression would find the area of this figure?

Mathematics
1 answer:
fgiga [73]3 years ago
3 0

Answer:

split the figure into two parts

1st part is 7 × (12-3)

2nd part is 3×4

(9×7) + (3×4)

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12. The white triangle drawn on the road sign in the picture has a height of 8" and a base of 5". Which formula would be correct
nevsk [136]

Step-by-step explanation:

I think you did not copy the correct information of your answer options here.

because, as you wrote the 4 options, none of them are correct.

remember, the area of a triangle is base times height divided by 2

A = (b × h) / 2

given the numbers in the problem statement

b = 5, h = 8

A = (5 × 8) / 2 = 40 / 2 = 20

you need to pick the answer option in your original document that would lead to 20.

none of the 4 options you gave me here would result in 20. so, there must be something wrong with them.

4 0
3 years ago
Help me out please thank you
N76 [4]

Well first you find the common denominatior of 1/7 and 4/9 which is 63 so then you see how many times does 7 go into 63 so this is part where you multiply 7 and 9 which gives you 63 and then you do how many times 9 go into 63 which is 7 so then you will take 7 times the 4 which is 28 so now your fractions should be 9/63 and 28/63 then now your next step is to add the 9 and 28 which is now 37/63 and you ask yourself and you divide that by any number to get a smaller number which is no so that is your answer

3 0
4 years ago
The length of a rectangle is one foot more than twice its width if the area of the rectangle is 300 ft^2 find the dimensions of
umka21 [38]
Let's say the length is L, and the width is W
L is one foot more than 2W, so L=1+2W
the area of the rectangle: A=L*W=300
replace L with 1+2W: (1+2W)*W=300 =>2W^2+W-300=0
solve the quadratic equation: (W-12)(2W+25)=0
W=12 or W=-12.5 (impossible)
so the width is 12, and the length is L=1+2*12=25

please refer to this website for how to factor a quadratic equation
http://www.purplemath.com/modules/factquad.htm

7 0
3 years ago
Read 2 more answers
Suppose a long jumper claims that her jump distance is less than 16 feet, on average. Several of her teammates do not believe he
Greeley [361]

Answer:

Claim : Jump distance is less than 16 feet.

A ) H_0:\mu \geq16\\H_a: \mu

n = 19

Since n <30 we will use t test

The mean distance of the sample jumps is 13.2 feet. i.e. x = 13.2 feet

The standard deviation of her jump distance is 1.5 ft i.e. s = 1.5

Formula : t =\frac{x-\mu}{\frac{s}{\sqrt{n}}}

Substitute the values:

B) t =\frac{13.2-16}{\frac{1.5}{\sqrt{19}}}

t =−8.136

degree of freedom df = n-1 = 19-1 = 18

\alpha = 0.1

t_{(\frac{\alpha}{2},df)}=1.73

t critical > t statistics

So, we accept the null hypothesis

C) So, the claim is wrong that ump distance is less than 16 feet.

Hence long jump distance must be greater than or equal to 16

5 0
3 years ago
The side lengths of a rectangular room are 10 ft and 24 ft. The room is going to be split in half along the hypotenuse of a
MA_775_DIABLO [31]

40ft hope I helped with your problem! Have a great day my comrade

6 0
3 years ago
Read 2 more answers
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