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3 years ago

In APQR, PR = 23 mm, QR = 39 mm, and m/R = 163º.

1 answer:

algol133 years ago

4 0

Use an inverse trigonometric ratio to find m∠A. You know the lengths of the ... In △PQR, PR = 23 mm, QR = 39 mm, and m∠R = 163°.

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Help Please Math!!! Find The Surface Area Of The Prism

LiRa [457]

Answer:

Surface Area=208 cm²

Step-by-step explanation:

Surface Area= 2×l×w+2×l×h+2×h×w

SA= 2×11×4+2×4×4+2×11×4=208

8 0

2 years ago

The range of the function f(x) = x + 5 is {7, 9). What is the function's domain?

Maru [420]

Answer:

(2,4)

Step-by-step explanation:

The range is the y values

y = x+5

Let y=7

7 =x+5

Subtract 5

7-5 x+5-5

2=x

Let y= 9

Subtract 5

9-5 x+5-5

4=x

The domain is (2,4)

8 0

3 years ago

Solve the equation -25p – 2 = 48 for p. A. -2 B. 1 C. 3 D. 8

konstantin123 [22]

Answer:

A. -2

Step-by-step explanation:

-25p - 2 = 48

First, add 2 to both sides

That gives you

-25p = 50

Then, divide both sides by -25. Then that leaves the answer.. p = -2

5 0

3 years ago

I need help on this question so if u actually can answer it correctly and show how you got it that would be helpful thx

Sedaia [141]

$\rm \looparrowright - \frac{1}{4} (8 - 4 + 20x)$

Subtract the numbers

$\rm \looparrowright - \frac{1}{4} (4+ 20x)$

Factor out 4 from the expression

$\rm \looparrowright - \frac{1}{ \cancel4} \times \cancel4(1 + 5x)$

$\rm \looparrowright - (1 + 5x)$

When there is a - in front of an expression in parentheses, change the sign of each term in the expression to -

$\rm \looparrowright - 1 - 5x$

Thus, Options A and F are correct!!~

6 0

2 years ago

In a missile-testing program, one random variable of interest is the distance between the point at which the missile lands and t

Arisa [49]

Answer:

Step-by-step explanation:

From the given data

we observed that the missile testing program

Y1 and Y2 are variable, they are also independent

We are aware that

$(Y_1)^2 and (Y_2)^2$ have $x^2$ distribution with 1 degree of freedom

and $V=(Y_1^2)+(Y_2)^2$ has x^2 with 2 degree of freedom

$F_v(v)=\frac{e^{-\frac{v}{2}}}2$

Since we have to find the density formula

$U=\sqrt{V}$

We use method of transformation

$h(V)=\sqrt{U}\\\\=U$

There inverse function is $h^-^1(U)=U^2$

We derivate the fuction above with respect to u

$\frac{d}{du} (h^-^1(u))=\frac{d}{du} (u^2)\\\\=2u^2^-^1\\\\=2u$

Therefore,

$F_v(u)=F_v(h-^1)(u)\frac{dh^-^1}{du} \\\\=\frac{e^-\frac{u^-^}{2} }{2} (2u)\\\\=e^-{\frac{u^2}{2} }U$

8 0

3 years ago