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3 years ago

Calculate the area of each shape below. Figures are not drawn to scale.

Mathematics

1 answer:

STatiana [176]3 years ago

6 0

Answer:

The area of the triangle is 168 meters

Step-by-step explanation:

Formula to find the area of a triangle:

Area = 1/2 Base x Height

So, first you need to identify the base and the height. The base is 16m + 8m, and the height is 14m. All you have to do is plug in the numbers:

Area = 1/2 (16 + 8) x 14

Then, you solve:

Area = 1/2 (16 + 8) x 14

Area = 1/2 24 x 14

Area = 1/2 336

Area = 168 meters

So, the area of the triangle is 168 meters

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(2x2 + 1) + 3/2x2=1=4

guajiro [1.7K]

Answer:

3(2×2+1)+3×(2×2)/3=1×3=4×3

36+3+4=3=12

36+3+4-3-12

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3 years ago

8-10-((-2) +(-5) - 3-(-4))

Ket [755]

Answer:

$8 - 10 - (( - 2) + ( - 5) - 3 - ( - 4))$

$8 - 10 - ( - 2 - 5 - 3 + 4)$

$8 - 10 - ( - 10 + 4)$

$8 - 10 - ( - 6)$

$8 - 10 + 6$

$14 - 10$

$4$

Hope it helps you :)

3 0

2 years ago

The “Limited Plan” offers a 15% discount for power usage between 7pm – 9am if most of your usage happens during that time. If no

Dovator [93]

The correct answer is:

$54.75.

Explanation:

We only receive the discount if the majority of our usage occurs between 7 pm to 9 am. However, in the month of March we used 450 KWH between these hours and 465 KWH between 9 am to 7 pm; this means we lose our discount.

The rate per KWH is $0.05432; this means we pay

450(0.05432) = 24.444 for the hours of 7 pm to 9 am.

We also pay $0.05432/kwh for the hours of 9 am to 7 pm, but we have an additional 20% premium for the usage between these hours; this means we pay

465(0.05432)(1.2) = 30.31056, for a total of

24.444+30.31056 = 54.75456 ≈ 54.75

4 0

3 years ago

Read 2 more answers

Lines m and n are parallel. They are cut by transversal t. What other angle is equal to 65 degrees?

Marizza181 [45]

Angle 3, they are opposite interior angles

5 0

4 years ago

Read 2 more answers

A rocket is shot straight up into the air with an initial velocity of 500 ft per second and from a height of 20 feet above the g

iragen [17]

The height of the rocket above the ground after $t$ seconds is given by the equation : $H= -16t^2+Vt+h$ , where $V$ is the initial velocity and $h$ is the initial height.

Given that, $V= 500 ft/second$ and $h= 20 ft$

So, the equation will become: $H= -16t^2 +500t+20$

A) For finding the height of the rocket 3 seconds after the launch, we will plug $t=3$ into the above equation. So....

$H= -16(3)^2+500(3)+20\\ \\ H= -16(9)+1500+20\\ \\ H= -144+1500+20=1376$

So, the height of the rocket 3 seconds after the launch is 1376 feet.

B) When the rocket at a height of 400 feet, then we will plug $H= 400$

$400=-16t^2+500t+20\\ \\ 16t^2-500t-20+400=0\\ \\ 16t^2-500t+380=0\\ \\ 4(4t^2-125t+95)=0\\ \\ 4t^2-125t+95=0$

Using quadratic formula, we will get......

$t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(95)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{14105}}{8}\\ \\ t= 30.4705... \\ and \\ t= 0.7794...$

So, after 0.7794...seconds and 30.4705...seconds the rocket is at a height of 400 feet above the ground.

C) The time duration that the rocket remains in the air means we need to find the time taken by the rocket to reach the ground. When it reaches the ground, then $H=0$ . So.....

$0=-16t^2+500t+20\\ \\ -4(4t^2-125t-5)=0\\ \\ 4t^2-125t-5=0$

Using quadratic formula, we will get.....

$t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(-5)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{15705}}{8}\\ \\ t=31.2899...\\ and\\ t= -0.0399...$

(Negative value is ignored as time can't be in negative)

So, the rocket will remain in the air for 31.2899... seconds.

5 0

4 years ago