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2 years ago

Drag each expression to show whether it is equivalent to −1.3−(−8.9) -1.3 x - -8.9 x , −1.3+(−8.9) -1.3 x + -8.9 x , or neither.

Mathematics

2 answers:

charle [14.2K]2 years ago

8 0

Answer:

1) 10.20x

2)7.60x

3)-10.20x

4)7.60x

5)7.60x

6)-7.60

scoundrel [369]2 years ago

3 0

I got one wrong but it goes on the neither spot

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A jumbo crayon is composed of a cylinder with a conical tip. The cylinder is 12 cm tall with a radius of 1.5 cm, and the cone ha

s2008m [1.1K]

Answer:

Part 1) The lateral area of the cone is $LA=2\pi\ cm^{2}$

Part 2) The lateral surface area of the cylinder is $LA=36\pi\ cm^{2}$

Part 3) The surface area of the crayon is $SA=41.50\pi\ cm^{2}$

Step-by-step explanation:

Part 1) Find the lateral area of the cone

The lateral area of the cone is equal to

$LA=\pi rl$

we have

$r=1\ cm$

$l=2\ cm$

substitute

$LA=\pi (1)(2)$

$LA=2\pi\ cm^{2}$

Part 2) Find the lateral surface area of the cylinder

The lateral area of the cylinder is equal to

$LA=2\pi rh$

we have

$r=1.5\ cm$

$h=12\ cm$

substitute

$LA=2\pi (1.5)(12)$

$LA=36\pi\ cm^{2}$

Part 3) Find the surface area of the crayon

The surface area of the crayon is equal to the lateral area of the cone, plus the lateral area of the cylinder, plus the top area of the cylinder plus the bottom base of the crayon

Find the area of the bottom base of the crayon

$A=\pi[r2^{2}-r1^{2}]$

where

r2 is the radius of the cylinder

r1 is the radius of the cone

substitute

$A=\pi[1.5^{2}-1^{2}]$

$A=1.25\pi\ cm^{2}$

Find the area of the top base of the cylinder

$A=\pi(1.5)^{2}=2.25\pi\ cm^{2}$

Find the surface area

$SA=2\pi+36\pi+2.25\pi+1.25\pi=41.50\pi\ cm^{2}$

8 0

2 years ago

Read 2 more answers

For the linear equation 5x - y = 2, which of the following has a value of 5?

Step2247 [10]

The slope-intercept form of a line is the following:

$y=mx+q$

In this form, $m$ is the shape and $q$ is the y-intercept. Your line written in this form is

$y=5x-2$

which implies that the slope is 5 and the y-intercept is -2.

8 0

3 years ago

Starting from rest the motor boat travels around the circular path (radius of curvature= 40 m),at a speed (v= 0.8 t m/s),where (

dsp73

Answer:

The magnitude of the boat's velocity is 5.66 m/s

Step-by-step explanation:

Given;

Radius of curvature of the circular path, r = 40 m

speed of the boat, v = 0.8 t m/s

The magnitude of the boat's velocity when it has traveled 20 m is ?

a = dv / dt,

v = 0.8 t (differentiate with respect to t)

dv / dt = 0.8 = a

thus, a = 0.8 m/s²

If the boat starts from rest, initial velocity, u = 0

Apply kinematic equation, to solve for the boat velocity at 20 m

v² = u² + 2as

v² = 0 + 2 x 0.8 x 20

v² = 0 + 32

v² = 32

v = √32

v = 5.66 m/s

Therefore, the magnitude of the boat's velocity when it has traveled 20 m is 5.66 m/s

7 0

2 years ago

The area of a square painting is 600 in.² to the nearest hundredth inch what is the perimeter of the painting?

Musya8 [376]

$\sqrt{600} = 24.49$
24.49*4

48.98

73.47

97.96

The perimeter of that square is 97.96.

6 0

3 years ago

A farmer outside of Clinton NC plants 20 plots of a new type of corn. The average yield for these plots is ī= 150 bushels per ac

bogdanovich [222]

Answer:

The 95% confidence interval is $(150\pm 8.8)$ .

Step-by-step explanation:

The information provided is:

$\bar x=150\\\sigma = 20\\n=20$

The critical value for 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the margin of error as follows:

$MOE=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}$

$=1.96\times\frac{20}{\sqrt{20}}\\\\=1.96\times 4.47214\\\\=8.7653944\\\\\approx 8.8$

Then the 95% confidence interval is:

$CI=\bar x\pm MOE$

$=150\pm 8.8$

Thus, the 95% confidence interval is $(150\pm 8.8)$ .

5 0

2 years ago