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Assoli18 [71]
3 years ago
6

Work out the value of 6x^2 when x= -2

Mathematics
2 answers:
natka813 [3]3 years ago
6 0
We are given the expression 6x² and a value of x, -2. Starting off, we need to plug in the value of x into 6x², which, when doing so, gives us 6(-2)². By the law of PEMDAS, we will need to do the exponent first. -2² is 4, and when we multiply 4 by 6, we get 24. Therefore, the value of 6x² when x = -2 is 24. Hope this helped and have a phenomenal day!
anzhelika [568]3 years ago
5 0
6(-2)^2
-18^2
-324
HOPE IT HELPS!!!

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KengaRu [80]

Answer:


Step-by-step explanation:

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5 0
3 years ago
Solve the following equation 5(y+4) =4(y+5)
kvv77 [185]

Answer:

Step-by-step explanation:

5(y+4) = 4(y+5)

5y+20 = 4y + 20

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6 0
3 years ago
Plz plz ans this with explanation
saveliy_v [14]

Answer:

8750

Step-by-step explanation:

Since the random sample contained 125 out of 500 people who voted for the dog park that is about 25%. since 25% of 500 is 125. If we scale that up to 25% of 3500 it would be 8750 people.

8 0
3 years ago
Read 2 more answers
The area of a rectangle is 66 ft2, and the length of the rectangle is 1 ft more than twice the width. Find the dimensions of the
umka21 [38]
This is probably way overcomplicated but i cant remember how to do it
w = n
l = 2n+1
w*l = 66
n(2n+1) = 66
2n^2 + n =66
2n^2 + n - 66 = 0
using quadratic equation

n = 11/2

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3 0
3 years ago
Help me i need to submit to my teacher by 8pm​
Vladimir [108]

Please see the figure. We'll first work out half the area of the rounded triangle, half the unshaded part, then double it, then subtract it from the big square.

Half the area is the circular sector PTQ (with center P, arc TQ) minus the right triangle PUT.

A/2 = area(sector PTQ) - area(triangle PUT)

The triangle is half of equilateral triangle PQT, so a 30/60/90 right triangle so we know the sides are in ratio 1:√3:2 so

TU = (7/2)√3

area(PUT) = (1/2) (7/2)(7/2)√3 = (49/8)√3

area(sector PTQ) = (angle TQP / 360°) πr^2

We know angle TQP is 60° because TQP is equilateral.  r=7.

area(sector PTQ) = (60°/360°) π (7²) = 49π/6

Putting it together,

A/2 = area(sector PTQ) - area(triangle PUT)

A = 2(49π/6 -  (49/8)√3)

A = 49(π/3 - √3/4) square cm

I hate ruining a nice exact answer with an approximation, but they seem to be asking.

A ≈ 30.095057615914535

Check:

I'm not sure how to check it.  I'd estimate it's about 25% bigger than equilateral triangle PQT with area (√3/4)7² ≈ 21.2, so around 27. 30 seems reasonable.

Now the real area we seek is the big square PQRS minus A, so

area = 7² - 30.095057615914535 = 18.904942384086 sq cm

They want square meters for some reason; we scale by (1/100)²

Answer: 0.00189 square meters

7 0
3 years ago
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