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3 years ago

QUESTION

Mathematics

2 answers:

Natalka [10]3 years ago

7 0

I agree with Karen!!!!!

garri49 [273]3 years ago

3 0

Answer:

Step-by step

What do we already know about the equation?

We know that they EACH bought a hockey puck($17) and 2 shirts

Half of 214 is 107. So they had to have spent 107 separately

So we make the equation and let s represent shirts

2s + 17 = 107

- 17 -17

2s= 90. DIVIDE by 2

s = 45

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Can someone confirm this? is it correct?

EastWind [94]

60 Sorry, but the value of 150 you entered is incorrect. So let's find the correct value. The first thing to do is determine how large the Jefferson High School parking lot was originally. You could do that by adding up the area of 3 regions. They would be a 75x300 ft rectangle, a 75x165 ft rectangle, and a 75x75 ft square. But I'm lazy and another way to calculate that area is take the area of the (300+75)x(165+75) ft square (the sum of the old parking lot plus the area covered by the school) and subtract 300x165 (the area of the school). So (300+75)x(165+75) - 300x165 = 375x240 - 300x165 = 90000 - 49500 = 40500 So the old parking lot covers 40500 square feet. Since we want to double the area, the area that we'll get from the expansion will also be 40500 square feet. So let's setup an equation for that: (375+x)(240+x)-90000 = 40500 The values of 375, 240, and 90000 were gotten from the length and width of the old area covered and one of the intermediate results we calculated when we figured out the area of the old parking lot. Let's expand the equation: (375+x)(240+x)-90000 = 40500 x^2 + 375x + 240x + 90000 - 90000 = 40500 x^2 + 615x = 40500 x^2 + 615x - 40500 = 0 Now we have a normal quadratic equation. Let's use the quadratic formula to find its roots. They are: -675 and 60. Obviously they didn't shrink the area by 675 feet in both dimensions, so we can toss that root out. And the value of 60 makes sense. So the old parking lot was expanded by 60 feet in both dimensions.

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3 years ago

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ANEK [815]

True
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2 years ago

Read 2 more answers

Simplify. cos^2[(pi/2) - x] + cos^2[(pi/2) - x] + cos^2x

Ipatiy [6.2K]

Numerator
cos(π/2−x)=cos(π/2)cosx+sin(π/2)sinx

now cos(π/2)=0 and sin(π/2)=1

simplifies to : 0 + sinx = sinx

Denominator

sin(π/2−x)=sin(π/2)cosx+cos(π/2)sinx

simplifies to : cosx + 0 = cosx

⇒cos(π/2−x)sin(π/2−x)=sinx/cosx=tanx

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3 years ago

Here is a triangular pyramid and its net.

bazaltina [42]

Answer:

a) Area of the base of the pyramid = $15.6\ mm^{2}$

b) Area of one lateral face = $24\ mm^{2}$

c) Lateral Surface Area = $72\ mm^{2}$

d) Total Surface Area = $87.6\ mm^{2}$

Step-by-step explanation:

We are given the following dimensions of the triangular pyramid:

Side of triangular base = 6mm

Height of triangular base = 5.2mm

Base of lateral face (triangular) = 6mm

Height of lateral face (triangular) = 8mm

a) To find Area of base of pyramid:

We know that it is a triangular pyramid and the base is a equilateral triangle. $\text{Area of triangle = } \dfrac{1}{2} \times \text{Base} \times \text{Height} ..... (1)\\$