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Evgesh-ka [11]
3 years ago
15

This figure is made up of a rectangle and

Mathematics
1 answer:
kap26 [50]3 years ago
3 0

768 is the correct answer to this problem

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What's the area of the quadrilateral ​
shepuryov [24]

Answer: Use the formula (base1+base2)/2 multiplied by height(3 in)

Step-by-step explanation: I can't see what the bottom base is, but lets say it is 12. 5+12=17, and 17/2=8.5. 8.5*3=25.5

Answer:25.5

5 0
2 years ago
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!Help !
svetoff [14.1K]

Answer:

50 minutes

Step-by-step explanation:

she spent 1 hr and 30 minutes on homework

she spends 2/3 of an hour on the phone = 40 minutes

1 hr 30 min + 40 minutes = 2 hrs and 10 min

so she had 50 minutes to kill before bedtime since she goes to sleep 3 hrs after dinner

2hr 10 min + 50 min = 3 hrs.

8 0
3 years ago
Find the circumference of each circle. Use 3.14 for the value of π . Round answer to the nearest tenth.
kvv77 [185]

Answer:

C=9.4\pi meters

Step-by-step explanation:

Recall to find the circumference of a circle the formula is C=\pi d or C=2\pi r.

We start by substituting 4.7m for r in the second circumference formula.

C=2\pi r\\C=2\pi (4.7)\\C=9.4\pi. We input this value of r into the area formula.


8 0
2 years ago
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How you would find the total area of the following composite shapes.
sergiy2304 [10]

Answer:

By breaking it up in shapes

Step-by-step explanation:

I annotated your photo and it's attached below :) that's an example on how to break shapes into easier parts. So yeah

Download pdf
7 0
2 years ago
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PLEASE HELP QUICK!! WILL OFFER 100 POINTS TO THE FIRST ONE THAT ANSWERS
Reil [10]

Given

x+1 = \sqrt{7x+15}

We have to set the restraint

x+1\geq 0 \iff x \geq -1

because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:

(x+1)^2=7x+15 \iff x^2+2x+1=7x+15 \iff x^2-5x-14 = 0

The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.

Similarly, we have

x-3 = \sqrt{x-1}+4 \iff x-7=\sqrt{x-1}

So, we have to impose

x-7\geq 0 \iff x \geq 7

Squaring both sides, we have

(x-7)^2=x-1 \iff x^2-14x+49=x-1 \iff x^2-15x+50 = 0

The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.

7 0
3 years ago
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