All categories

3 years ago

-50 + 5/13p when p= -26

Mathematics

2 answers:

kaheart [24]3 years ago

7 0

Answer:

-60

Step-by-step explanation:

Marysya12 [62]3 years ago

3 0

16905/338 hope it helps

You might be interested in

Which point below is a solution to the equation y = 2x + 3

pashok25 [27]

Answer:

The x-intercept it (-1.5, 0) and the y-intercept is (0, 3). The slope is 2.

Step-by-step explanation:

7 0

3 years ago

What is 9 divided by 3/4

Valentin [98]

If you would like to know what is 9 divided by 3/4, you can calculate this using the following steps:

9 / 3/4 = 9 * 4/3 = 12

The correct result would be 12.

4 0

4 years ago

Read 2 more answers

Combine the like terms to create an equivalent expression 2q+2+9

Katyanochek1 [597]

Answer:

2q + 11

Step-by-step explanation:

Add like terms.

2q + 2 + 9

2 and 9 are like terms because they don't have letters like 2q.

2q + (2 + 9)

2q + 11

7 0

3 years ago

Suppose that X is a binomial random variable with n = 10 and p = 0.25. Find P 2).

goblinko [34]

Answer:

0.3582 0.3600 x is the binomial random variable

3 0

3 years ago

Read 2 more answers

Please find the perimeter and area of these shapes

ladessa [460]

Answer:

ABC shaded area = 36 $\pi$ - 72 cm²

ABC shaded area perimeter = $6\pi +12\sqrt{2}$ cm

ABCD area = $\dfrac52 \pi$ cm²

ABCD perimeter = $3\pi +2$ cm

Step-by-step explanation:

Shape ABC

Assuming you want the area and perimeter of the shaded part of the shape only...

Area

Area of a sector = $\dfrac12r^2\theta$ (where r is the radius and $\theta$

⇒ area of a sector = $\dfrac12 \times 12^2\times \dfrac{\pi}{2} =36\pi \ \textsf{cm}^2$

Area of triangle = 1/2 x base x height

⇒ area of triangle = 1/2 x 12 x 12 = 72 cm²

Therefore, area of shaded area = area of sector - area of triangle

⇒ area = 36 $\pi$ - 72 cm²

Perimeter

Arc length = $r\theta$ (where r is the radius and $\theta$

⇒ arc length = $12\times\dfrac12\pi =6\pi \ \textsf{cm}$

Hypotenuse of triangle = $\sqrt{a^2+b^2}$ (where a and b are the legs of the right triangle)

⇒ hypotenuse = $\sqrt{12^2+12^2} =12\sqrt{2}$ cm

Therefore, perimeter = arc length + hypotenuse

⇒ perimeter = $6\pi +12\sqrt{2}$ cm

Shape ABCD

Area

Area of a semicircle = $\dfrac12 \pi r^2$ (where r is the radius)

⇒ area of large semicircle ABC = $\dfrac12 \times \pi \times 2^2=2\pi \ \textsf{cm}^2$

⇒ area of small semicircle AD = $\dfrac12 \times \pi \times 1^2=\dfrac12\pi \ \textsf{cm}^2$

⇒ area of shape ABCD = $\dfrac12 \pi + 2 \pi=\dfrac52 \pi \ \textsf{cm}^2$

Perimeter

1/2 circumference = $\pi r$

⇒ perimeter = $2\pi +2+\pi=3 \pi+2 \ \textsf{cm}$

7 0

2 years ago