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3 years ago

The perimeter of a rectangular garden is 302 m.

Mathematics

1 answer:

kompoz [17]3 years ago

4 0

I think that person above me should be your answer

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I need the answer cause this is for a grade please!!! Jamie has a rectangular room in her home with a width of 8 feet and a diag

FinnZ [79.3K]

Answer:

You can find the diagonal of a rectangle if you have the width and the height. The diagonal equals the square root of the width squared plus the height squared.

92+72−−−−−−√=11.4

Step-by-step explanation:

4 0

3 years ago

Nolan just drove at a constant rate for 5 hours. He is now 340 miles from where he started. At what rate was he driving? Nolan h

Sauron [17]

Answer:

68 miles

3 hours

Step-by-step explanation:

340/5=68

204/68=3

7 0

2 years ago

In a video game, a bird is flung on a sling shot and follows a parabolic path. The path of the bird can be modeled by the equati

Novay_Z [31]

Answer:

The bird will take 0.88 seconds to reach a height of 40 feet in the air for the first time

Step-by-step explanation:

we have

$h=-16t^2+55t+4$

This is a vertical parabola open downward

The vertex represent a maximum

where

h is the height of the bird

t is the amount of time, in seconds, the bird had been in the air

For h=40 ft

substitute in the quadratic equation

$40=-16t^2+55t+4$

$-16t^2+55t+4-40=0$

$-16t^2+55t-36=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-16t^2+55t-36=0$

$a=-16\\b=55\\c=-36$

substitute in the formula

$x=\frac{-55\pm\sqrt{55^{2}-4(-16)(-36)}} {2(-16)}$

$x=\frac{-55\pm\sqrt{721}} {-32}$

$x=\frac{-55+\sqrt{721}} {-32}=0.88\ sec$

$x=\frac{-55-\sqrt{721}} {-32}=2.56\ sec$

therefore

The bird will take 0.88 seconds to reach a height of 40 feet in the air for the first time

3 0

3 years ago

A rectangle has an area of 126 m2 and a width of 7 m. What is its length?

Lorico [155]

126 divided by 7 equals 18

18 m

8 0

3 years ago

Read 2 more answers

Sin α = 21/29, α lies in quadrant II, and cos β = 15/17, β lies in quadrant I Find sin (α - β).

Sever21 [200]

$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$

$\sin\alpha=\dfrac{21}{29}\implies \cos^2\alpha=1-\sin^2\alpha=\dfrac{400}{841}$

Since $\alpha$ lies in quadrant II, we have $\cos\alpha$ , so

$\cos\alpha=-\sqrt{\dfrac{400}{841}}=-\dfrac{20}{29}$

$\cos\beta=\dfrac{15}{17}\implies\sin^2\beta=1-\cos^2\beta=\dfrac{64}{289}$

$\beta$ lies in quadrant I, so $\sin\beta>0$ and

$\sin\beta=\sqrt{\dfrac{64}{289}}=\dfrac8{17}$

So

$\sin(\alpha-\beta)=\dfrac{21}{29}\dfrac{15}{17}-\left(-\dfrac{20}{29}\right)\dfrac8{17}=\dfrac{475}{493}$

8 0

3 years ago