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3 years ago

13

She was so famous she had 12 body guards at one concert. At the next concert she had 15 body guards. What is the percent of chan

ge in the number of body guards

1 answer:

balu736 [363]3 years ago

3 0

Answer:

she increased her body guards by 20%

Step-by-step explanation:

15-12 = 3

3/15 = .2 or 20%

OR you can do it like...12/15 = .8

100%-80% = 20% more body guards.

hope that helps

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Compare the two expressions

Art [367]

Yes, they are equivalent
6+7-5=8. -5+10+3=8

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4 years ago

Graph a triangle (STU) and reflect it over the y-axis to create triangle ST'U'.

lukranit [14]

The x-coordinates of $\triangle S'T'U'$ will be the negation of the x-coordinates of $\triangle STU$

The line segment from S to the y-axis equals the line segment from S' to the y-axis. Similarly, the line segment from T to the y-axis equals the line segment from T' to the y-axis

See attachment for $\triangle STU$ and $\triangle S'T'U'$

In order to solve this question, I will make the following assumptions.

Assume that the coordinates of $\triangle STU$ are

$S = (4,5)$

$T = (5,9)$

$U=(3,8)$

Refer to attachment for illustrations

<u>(1) Reflect </u> $\triangle STU$ <u> over y-axis and describe the transformation</u>

To reflect $\triangle STU$ across the y-axis, the following rule must be followed

$(x,y) \to (-x,y)$

This means that:

$S = (4,5) \to S' = (-4,5)$

$T = (5,9) \to T' = (-5,9)$

$U=(3,8) \to U'=(-3,8)$

<u>The description of the </u><u>transformation </u><u>is as follows:</u>

Notice that the signs of the x-coordinates $\triangle STU$ and $\triangle S'T'U'$ of both triangles are different.

In other words, if the x-coordinate of one is positive, then the other will have a negative x-coordinate; and vice versa.

<u>(2) Compare the segments and the line of reflection</u>

To reflect across the y-axis means that the reflecting line is the y-axis, itself.

The distance between a point to the y-axis is the absolute value of the x-coordinate.

So, the distance between S and the y-axis is:

$S = |4| = 4$

The distance between S' and the y-axis is:

$S' = |-4| = 4$

We can conclude that the two line segments are equal.

This is the same for other point T and T' because of the formula used above.

<u>From T and T' to the y-axis is:</u>

$T =|5| =5$

$T' =|-5| =5$

Read more at:

brainly.com/question/938117

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3 years ago

1. Y= 3x + 5<br> +0<br> 2<br> Increasing<br> Decreasing<br> Positive<br> Negative

Natalka [10]

Answer:

The derivative of a function may be used to determine whether the function is increasing or decreasing on any intervals in its domain. If f′(x) > 0 at each point in an interval I, then the function is said to be increasing on I. f′(x) < 0 at each point in an interval I, then the function is said to be decreasing on I.

Step-by-step explanation:

7 0

3 years ago

Mrs. R is pregnant and very worried that the child will have beta-thalassemia. What is the probability that the child will have

kenny6666 [7]

Answer:

There are three kind of beta-thalassemia.

Total possible outcome= 3 types of beta- thalassemia + 1 [child will have no trace of disease] = 4

As we know Probability(outcome) = $\frac{\text{ Total favorable outcome}}\text{Total possible outcome}$

Probability that the child will have beta-thalassemia major =\frac{_{1}^{3}\textrm{C}}{_{1}^{4}\textrm{C}}

= $\frac{3}{4}$

∴ favorable= $_{1}^{3}\textrm{C}$

And , Total = $_{1}^{4}\textrm{C}$

Probability that the child will have no trace of the disease= $\frac{1}{4}$

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4 years ago

In the figure above, PQRS is a circle. If PQT and SRT<br>are straight lines, find the value of x.

Elena L [17]

Given:

PQRS is a circle, PQT and SRT are straight lines.

To find:

The value of x.

Solution:

Since PQRS is a circle, PQT and SRT are straight lines, therefore, PQRS isa cyclic quadrilateral.

We know that, sum of opposite angles of a cyclic quadrilateral is 180 degrees.

$m\angle SPQ+m\angle QRS=180^\circ$

$81^\circ+m\angle QRS=180^\circ$

$m\angle QRS=180^\circ-81^\circ$

$m\angle QRS=99^\circ$

Now, SRT is a straight line.

$m\angle QRT+m\angle QRS=180^\circ$ (Linear pair)

$m\angle QRT+99^\circ=180^\circ$

$m\angle QRT=180^\circ-99^\circ$

$m\angle QRT=81^\circ$ ...(i)

According to the Exterior angle theorem, in a triangle the measure of an exterior angle is equal the sum of the opposite interior angles.

Using exterior angle theorem in triangle QRT, we get

$m\angle PQR=m\angle QRT+m\angle QTR$

$x=81^\circ+22^\circ$

$x=103^\circ$

Therefore, the value of x is 103 degrees.

4 0

3 years ago