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4 years ago

Someone do this please cause i fell asleep in class

2 answers:

8 0

Gas in July is 2.3 times Gas in June
That means Gas in July (252.57) equals 2.3 times gas in June, or “n”
242.57= 2.3 x n and solve for n
Divide both sides by2.3
Calculator gives 105.465
Rounded to nearest HUNDRED is 100

Rounded to nearest HUNDREDTH would be 105.47
Check the answer to see if July gas is 2.3 times June gas
June gas 105.47 x 2.3 is 242.57
So the answer is 105.47 Rounded to nearest hundredth

Norma-Jean [14]4 years ago

7 0

Answer:

Answer is 105.47

Step-by-step explanation:

I did this in a exit ticket today are you from k12?

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You randomly draw A marble from a bag, record it’s color, and then replace it. You draw a blue marble 11 out of 50 times. What i

BlackZzzverrR [31]

Answer:

since you recorded the first marble then replaced it, you drew 10 blue marbles, so therefore we can predict the next marble you drew, was blue.

Step-by-step explanation:

4 0

3 years ago

True or False

kifflom [539]

Answer:

true
true
false
false

Step-by-step explanation:

One characteristic of a reflection that is useful for answering this question is that it always reverses the clockwise/counterclockwise orientation of a figure. Rotation and translation have no effect on that orientation.

1. translation by reflection

Reflection across two parallel lines has the net effect of translating a figure twice the distance between the parallel lines. So, one way to effect a translation using two lines of reflection is to draw one of them through the perpendicular bisector of a point and its translated image. Then the other line would be drawn parallel to the first through the image point.

True: translation can be replaced by two reflections

2. translation by rotation

A single point can be moved from one set of coordinates to another by rotating around any point on the perpendicular bisector of the original and its image. To "undo" the change in direction of other points in the image, the image can be rotated an equal angle in the reverse direction about the point that is in its proper place.

That is, if we rotate figure ABC an amount of X° about a point on the perpendicular bisector of AA', so that A ends up at A', then the translation can be finished by rotating that figure by -X° about point A'.

The simplest case is an initial rotation of 180° about the midpoint of AA', followed by another rotation of 180° about A'.

True: translation can be replaced by two rotations

3. rotation by reflection

As discussed above, reflection changes orientation and rotation does not.

However, a rotation can be replaced by two reflections. The rotation angle is equal to twice the angle between the lines of reflection. The point where the lines of reflection meet is the center of rotation.

False: rotation can be replaced by reflection

4. reflection by rotation and translation

As discussed above, reflection changes orientation, but rotation and translation do not.

False: reflection can be replaced by rotation and translation

6 0

3 years ago

Need ASAP plz help!! These are the numbers you use. 15 27 28 31 34 42 52

Dahasolnce [82]

im sorry but did you ever get the answer?

6 0

3 years ago

How much is the percent change if the price was $75 and now it is $300?

ale4655 [162]

Answer:

300 percent change

Step-by-step explanation:

300 is 400 percent of 75.(75x4 is 300.) Therefore, the percent change is

400 - 100 = 300.

6 0

3 years ago

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$tan(\alpha)=cot(\beta)$

Find the value of the length side adjacent to the angle alpha

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

6 0

3 years ago