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Pachacha [2.7K]
3 years ago
15

Find the area of the shape provided

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
7 0

Answer:

111m^{2}

Step-by-step explanation:

First you need to break the shape up to rectangles to make it easier to solve.

bottom rectangle : 15 x 3 = 45m^2

Top rectangle : 12 x 3=36m^2

Middle rectangle : 12-3-3=6 (because we already solved the top and bottom part of it)

we need to work out the base so we do 15-5= 5cm

6 x 5 = 30m^2

add them all together

30+36+45=111m^2

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Find the values of A B C AND D of <br> 4x^2 (2x^3+5x)= Ax^B +Cx^D
iren2701 [21]

The values are A=8, B=5, C= 20 and D=3

Explanation:

The expression is 4x^{2} (2x^{3} +5x)=Ax^{B} +Cx^{D}

Simplifying, we get,

8x^{5} +20x^3=Ax^{B} +Cx^{D}

Since, both sides of the expression are equal, we can equate the corresponding values of A, B, C and D.

Thus, we get,

8 x^{5}=A x^{B} ⇒ A=8 and B=5

Also, equating, 20 x^{3}=C x^{D}, we get,

C=20 and D=3

Thus, the values are A=8, B=5, C= 20 and D=3

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You purchase Aquasonic Gel for the ultrasound labs in the hospital. A 5-liter bottle costs $20.00. A case (12 tubes/case) of 250
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Large bottle....$37.5
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On a quiz, you answered 38 questions correctly out of 43. What percent did you answer correctly? (Round your answer to nearest p
horsena [70]

Answer:

D

Step-by-step explanation:

<u>38=x</u>

43=100

Cross multiply.

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1.) What is the equation of the path of firework #1? Write your equation in general form.
valina [46]

1. I'm assumig that the paths are perfect parabolas

this means that their general forms can be written in y=ax^2+bx+c

it's easier to find vertex form first then expand to get general form

vertex form is y=a(x-h)^2+k where the vertex is (h,k) and a is a constant


firework #1

vertex is (10,50), so (h,k)=(10,50) and h=10, k=50

h_1=a(t-10)^2+50

to find the value of a, subsitute another point

(0,0)

0=a(0-10)^2+50)

0=100a+50

a=\frac{-1}{2}

so the equation in vertex form is h_1=\frac{-1}{2}(t-10)^2+50

expand to get general form

h_1=\frac{-1}{2}(t^2-20t+100)+50

h_1=\frac{-1}{2}t^2+10t-50+50

h_1=\frac{-1}{2}t^2+10t





2.

same as last time

vertex is (10,72) so (h,k)=(10,72) so h=10 and k=72

equation is h_2=a(t-10)^2+72

find a

use another point

(0,22)

22=a(0-10)^2+72

22=100a+72

-50=100a

a=\frac{-1}{2}

so the equation in vertex form is h_2=\frac{-1}{2}(t-10)^2+72




3.

range is the numbers that h is allowed to be

think about what h represents. it represents the height of the rocket

from the graph, we can see that the lowest possible height is 0yd and the highest height is 50yd

so range is 0 to 50 or 0≤h≤50


domain is the numbers that t is allowed to be

think about what t represents. it represents how long the rocket has been flying

it will stop flying when it hits the ground or at t=20

it starts flying at t=0

so domain is from 0 to 20 or 0≤t≤20

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3 years ago
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