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3 years ago

What is the area of the parallelogram? please answer asap I'm begging!

Mathematics

2 answers:

DENIUS [597]3 years ago

8 0

Answer:

112 in²

Step-by-step explanation:

Area of a parallelogram:

Base × height

14 × 8

112 in²

nataly862011 [7]3 years ago

7 0

Answer:

A =112 in^2

Step-by-step explanation:

The area of a parallelogram is found by

A = bh where b is the base length and h is the height

A = 14*8

A =112 in^2

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2 divided by 41.000 show all work

charle [14.2K]

2 in 41.000
So ... 2 into 4 go's twice so put down 2
2 go's into 1 zero so put zero down. Then ur point
Ur real answer is 20.5
Hope that helped

8 0

4 years ago

I’m not sure where to begin

Dmitry [639]

(a)

we are given

the price of apple suddenly up by $0.75

Let's assume original price be 'p'

we are given

Sam bought 3 pounds of apple at the new price total is $5.88

so, new price = old price +0.75

price of 3 apples $=3(p+0.75)$

we are given that price is 5.88

so, we get equation as

$5.88=3(p+0.75)$ .......Answer

(b)

Since, original price is p

so, we have to solve for p

$5.88=3p+2.25$

$5.88-2.25=3p+2.25-2.25$

$3.63=3p$

$p=1.21$

so, original price is $1.21 per pound........Answer

8 0

3 years ago

Evaluate s(t)=∫t−[infinity]||r′(u)||du for the bernoulli spiral r(t)=⟨etcos(8t),etsin(8t)⟩. It is convenient to take −[infinity]

sveta [45]

$\vec r(t)=\langle e^t\cos8t,e^t\sin8t\rangle$

$\|\vec r'(t)\|=\sqrt{(e^t(\cos8t-8\sin8t))^2+(e^t(\sin8t+8\cos8t))^2}=e^t\sqrt{(\cos8t-8\sin8t)^2+(\sin8t+8\cos8t)^2}$

$\implies\|\vec r'(t)\|=e^t\sqrt{65}$

Then

$s(t)=\displaystyle\sqrt{65}\int_{-\infty}^te^u\,\mathrm du=\sqrt{65}e^t$

3 0

3 years ago

How do you simplify the radical √144 ?

SVETLANKA909090 [29]

Answer:

its 12

Step-by-step explanation:

you find whats closest for example, 10 squared is 100.

100 is close to 144

and 11x11= 121

so you would do 12

12x12 is 144

8 0

3 years ago

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Can someone tell me what i did wrong cuz it’s wrong please will name brainliest

valkas [14]

Switch 2 units to 5 units.

Explanation:

If Figure G was a reflection over the x-axis (which it is), then the bottom right angle would be at 4 on the y axis. In order for it to get to -1, you would need to move Figure G down 5 units. I hope this helps!

4 0

2 years ago

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