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1 year ago

explain how the graph of y=-2sin(x)-1 differs from the parent function. Make sure to compare key points used for graphing

Mathematics

1 answer:

Anna35 [415]1 year ago

5 0

You can obtain the graph of the function y=-2sinx-1 from the graph of the function y=sinx making such transformations:
1. the multiplication by -2 increases y value twice and then symmetrically reflects the graph y=2sinx about the x-axis;
2. adding -1 to the y=-2sinx you translate the graph 1 unit down.

On the added picture black line is a graph of y=sinx and red line is a graph of y=-2sinx-1.

Check:
1. When x=0, y=-2sin0-1=-1;
2. When $x= \frac{3\pi}{2}$ , then $y=-2\sin \frac{3\pi}{2} -1=-2(-1)-1=2-1=1$ .

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A satellite in a 100-mile-high circular orbit around Earth has a velocity of approximately 17,500 miles per hour.If this velocit

AveGali [126]

Answer:

64.3 miles ( D)

Step-by-step explanation:

Circular orbit height ( from the center of the earth ) = 100 mile

angle between satellite and earth = 50°

escape velocity = √2 * 17500 = 24748.73 miles /hour

distance between surface of earth and satellite

x = cos ∅

∴ Cos ∅ = x / 100

x = cos 50° * 100 ≈ 64.3 miles

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2 years ago

Peter wants to cut a rectangle of size 6x7 into squares with integer sides. What is the smallest number of squares he can get

Semmy [17]

He can cut 5 squares, by making 1 4 × 4, 2 3 × 3 and 2 2 × 2 squares.

5 0

2 years ago

An experiment was performed to compare the wear of two different laminated materials. Twelve pieces of material 1 were tested by

GenaCL600 [577]

Answer:

At 0.05 level of significance, the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Step-by-step explanation:

We hypothesize that mean difference between abrasive wear of material 1 and material 2 is greater than 2.

So we write the null hypothesis $H_0 : \mu_1 - \mu_2 >2$ ,

and the alternative hypothesis $H_1: \mu_1 - \mu_2 \leq 2$ .

We will find the T-score as well as the p-value. If the p-value is less than the level of significance, we will reject the null hypothesis, i.e. we will conclude that the abrasive wear of material 1 is less than that of material 2. Otherwise, we will accept the null hypothesis.

Since the variance is unknown and assumed to be equal, we will use the pooled variance

$s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2} = 20.05$ ,

where $n_1 = 12, n_2 =10, s_1 =4, s_2 = 5$ .

The mean of material 1 and material 2 are $\mu_1 =85, \mu_2=81$ respectively and mean difference $d$ is equal to 4. The hypothesize difference $d_0$ is equal to 2.

To find the T-score, we use the following formula

$T = \frac{d - d_0}{\sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2} }}$

Substituting all the values into the T-score formula gives us $T = 1.04$ , and the respective p-value is equal to 0.31. This means we have enough statistical evidence not to reject the null hypothesis, and at 5% significance level, the abrasive wear of material 1 exceeds that of material 2 by more than 2 units.

6 0

1 year ago

The box which measures 70cm X 36cm X 12cm is to be covered by a canvas. How many meters of canvas of width 80cm would be require

grigory [225]

Answer:

142.2 meters.

Step-by-step explanation:

We have been given that a box measures 70 cm X 36 cm X 12 cm is to be covered by a canvas.

Let us find total surface area of box using surface area formula of cuboid.

$\text{Total surface area of cuboid}=2(lb+bh+hl)$ , where,

$l$ = Length of cuboid,

$b$ = Breadth of cuboid,

$w$ = Width of cuboid.

$\text{Total surface area of box}=2(70\cdot36+36\cdot 12+12\cdot 70)$

$\text{Total surface area of box}=2(2520+432+840)$

$\text{Total surface area of box}=2(3792)$

$\text{Total surface area of box}=7584$

Therefore, the total surface area of box will be 7584 square cm.

To find the length of canvas that will cover 150 boxes, we will divide total surface area of 150 such boxes by width of canvass as total surface area of canvas will also be the same.

$\text{Width of canvas* Length of canvass}=\text{Total surface area of 150 boxes}$

$80\text{ cm}\times\text{ Length of canvass}=150\times 7584\text{cm}^2$

$\text{ Length of canvass}=\frac{150\times 7584\text{ cm}^2}{80\text{ cm}}$

$\text{ Length of canvass}=\frac{1137600\text{ cm}^2}{80\text{ cm}}$

$\text{ Length of canvass}=14220\text{ cm}$

Let us convert the length of canvas into meters by dividing 14220 by 100 as 1 meter equals to 100 cm.

$\text{ Length of canvass}=\frac{14220\text{ cm}}{100\frac{cm}{m}}$

$\text{ Length of canvass}=\frac{14220\text{ cm}}{100}\times\frac{m}{cm}$

$\text{ Length of canvass}=142.20\text{ m}$

Therefore, 142.2 meters of canvas of width 80 cm required to cover 150 such boxes.

5 0

2 years ago

On this graph, 4:00 p.m. occurs at 16 hours after midnight, and 6:00 p.m. occurs at 18 hours after midnight. Which statements ar

KatRina [158]

Answer:

TRUE OPTIONS ARE:

"The temperature increased until 4:00 p.m. "

"The temperature decreased after 6:00 p.m. "

"The temperature increased more quickly between 12:00 p.m. and 4:00 p.m. than before 12:00 p.m."

Step-by-step explanation:

graph attached and complete question below:

Which statements are true about the temperatures Luis recorded on the graph? Check all that apply.

The temperature increased until 4:00 p.m.
The temperature was not recorded between 4:00 p.m. and 6:00 p.m.
The temperature decreased after 6:00 p.m.
The temperature increased and then decreased before holding constant.
The temperature increased more quickly between 12:00 p.m. and 4:00 p.m. than before 12:00 p.m.

Until 1600 hours, the graph increases, so we can say temperature increased until 4.00 pm (FIRST OPTION TRUE).

From 1600 to 1800 hours (4 - 6pm), the temperature stayed same (horizontal line). This doesn't mean the temperature wasn't recorded. (2nd OPTION FALSE).

After 1800 hours (6pm), the line goes downward, so temperature decreased after 6 pm. (3rd OPTION TRUE).

If you look at the temperature graph, we can see temperature increased, then increased more, then constant, then decreased. Thus the 4th option isnt true. (4th OPTION FALSE).

Before 12, the increase isn't as sharp as after 12. After 12 temperature increase has more slope, so this increase is more. (5th OPTION TRUE).

7 0

1 year ago

Read 2 more answers