Target due

Caiden wants to use the distributive property to find the value of 28⋅19+28⋅6528·19+28·65. Which expression can he use?

MakcuM [25]

28 times 28 times (19+65)

4 0

4 years ago

The figure shows a cylinder of diameter 12cm and height = 15cm. A hole in the shape of cone is bored into one of its end. If the

Lisa [10]

Answer:

$\bold{495\pi} \approx \bold{1555.088 cm^3}$

Step-by-step explanation:

There was no figure but the question is clear

Volume of a cylinder is given by the formula $\bold{\pi r^2h}\\$

where r is radius of base of cylinder, h is the height

Volume of a cone is given by $\bold{\frac{1}{3} \pi r^2 h}$

where r is the radius of base of cone, h is the height

The radius of the cylinder = $\frac{1}{2}$ (diameter) = $\frac{1}{2}$ (12) = 6cm

Height of cylinder = 15cm

Volume of cylinder $V_{cyl} = \pi (6)^2 15 = \pi (36)15 = \bold{540\pi}$

Radius of cone = $\frac{1}{2}$ (radius of cylinder) = $\frac{1}{2}$ (6) = 3 cm

Height of cone same as height of cylinder = 15cm

Volume of cone, $V_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3)^2 15 = \frac{1}{3}(9)15\pi = \bold{45\pi}\\$

Difference is the volume of the remaining solid

$V_{cyl} - V_{cone} = 540\pi - 45\pi = \bold{495\pi} \approx \bold{1555.088 cm^3}$

5 0

2 years ago

What is 42/70 x 20/22

rusak2 [61]

Answer:

6/11 or 0.545

6 0

3 years ago

In ΔABC, point M is the midpoint of side AB and point D is the midpoint of segment MC. Prove that the area of ΔADC= the area of

Anna11 [10]

A line segment from a vertex to the midpoint of the opposite side is a "median". A median divides the area of the triangle in half, as it divides the base in half without changing the altitude.
AAMC is half AABC. AADC is half AAMC, so is 1/4 of AABC. (By the formula for area of a triangle.)
ABMC is half AABC. ABMD is half ABMC, so is 1/4 of AABC. (By the formula for area of a triangle.)
Then, AADC = 1/4 AABC = ABMC, so AADC = ABMC by the transitive property of equality.

8 0

3 years ago

Mhanifa can you please help? I will mark brainliest!

Rzqust [24]

Answer:

Given function

h(x) = 2x - 1

#15 Find the inverse of h(x)

Substitute x with y and h(x) with x and solve for y:

x = 2y - 1
2y = x + 1
y = 1/2x + 1/2

The inverse is:

h⁻¹(x) = 1/2x + 1/2

#16 The graph with both lines is attached.

The x- and y-intercepts of both functions have reversed values.

#17 Table of the inverse function will contain same numbers with swapped domain and range.

Initial look is like this:

x | -3 | -2 | -1 | 0 | 1 | 2 | 3
h⁻¹(x) | -1 | | 0 | | 1 | | 2

The rest of the table is filled in by finding the values:

x | -3 | -2 | -1 | 0 | 1 | 2 | 3
h⁻¹(x) | -1 | -0.5 | 0 | 0.5 | 1 | 1.5 | 2

7 0

3 years ago