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3 years ago

Plz help me with my math

Mathematics

1 answer:

rodikova [14]3 years ago

8 0

Answer: The answer is C

Step-by-step explanation:

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beth is making the minimum wage of $7.25 per hour. her boss promises her a raise of 5% how much more per week will she make if s

Korvikt [17]

Answer:

$13.60 difference approximately.

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3 years ago

What is the estimate of 137,638 plus 52,091

polet [3.4K]

The answer is 189,729
If to be rounded answer would be 190,000 I would say best estimate would be 180,000

3 0

3 years ago

Read 2 more answers

Please help and show how you got the answer:

kari74 [83]

4 - 6n = 60 - 2n subtract 4 from both sides

-6n = 56 - 2n add 2n to both sides

-4n = 56 divide both sides by (-4)

n = - 14

-7 + 11p = 3p - 47 add 7 to both sides

11p = 3p - 40 subtract 3p from both sides

8p = -40 divide both sides by 8

p = - 5

3a - 28 - 7a = 10a combine like terms

(3a - 7a) - 28 = 10a

-4a - 28 = 10a add 28 to both sides

-4a = 10a + 28 subtract 10 from both sides

-14a = 28 divide both sides by (-14)

a = - 2

17 - 5r + 9r = 12 + 6r - 1 combine like terms

17 + (-5r + 9r) = (12 - 1) + 6r

17 + 4r = 11 + 6r subtract 17 from both sides

4r = -6 + 6r subtract 6r from both sides

-2r = -6 divide both sides by (-2)

r = 3

8(y - 7) = -2(y + 3) use distributive property: a(b + c) = ab + ac

(8)(y) + (8)(-7) = (-2)(y) + (-2)(3)

8y - 56 = -2y - 6 add 56 to both sides

8y = -2y + 50 add 2y to both sides

10y = 50 divide both sides by 10

y = 5

-3(8k + 5) = 3(9 - k) use distributive property: a(b + c) = ab + ac

(-3)(8k) + (-3)(5) = (3)(9) + (3)(-k)

-24k - 15 = 27 - 3k add 15 to both sides

-24k = 42 - 3k add 3k to both sides

-21k = 42 divide both sides by (-21)

k = -2

2(3v - 5) = 2(v - 11) - 4 use distributive property: a(b + c) = ab + ac

(2)(3v) + (2)(-5) = (2)(v) + (2)(-11) - 4

6v - 10 = 2v - 22 - 4

6v - 10 = 2v - 26 add 10 to both sides

6v = 2v - 16 subtract 2v from both sides

4v = -16 divide both sides by 4

v = -4

-2/3 (15x + 3) -3x - 9 use distributive property: a(b + c) = ab + ac

= (-2/3)(15x) + (-2/3)(3) - 3x - 9

= (-2)(5x) - 2 - 3x - 9

= -10x - 2 - 3x - 9 combine like terms

= (-10x - 3x) + (-2 - 9)

= -13x - 11

3(1 - 9a) + 22a = 2(2a - 9) - 15

use distributive property: a(b + c) = ab + ac

(3)(1) + (3)(-9a) + 22a = (2)(2a) + (2)(-9) - 15

3 - 27a + 22a = 4a - 18 - 15 combine like terms

3 + (-27a + 22a) = 4a + (-18 - 15)

3 - 5a = 4a - 33 subtract 3 from both sides

-5a = 4a - 36 subtract 4a from both sides

-9a = -36 divide both sides by (-9)

a = 4

-3(3m - 10) - 7 = -5(m + 1) + 3m

use distributive property: a(b + c) = ab + ac

(-3)(3m) + (-3)(-10) - 7 = (-5)(m) + (-5)(1) + 3m

-9m + 30 - 7 = -5m - 5 + 3m combine like terms

-9m + (30 - 7) = (-5m + 3m) - 5

-9m + 23 = -2m - 5 subtract 23 from both sides

-9m = -2m - 28 add 2m to both sides

-7m = -28 divide both sides by (-7)

m = 4

6 0

3 years ago

Can anyone tell me what the answer please

Tasya [4]

What you're looking at in this image are two alternate exterior angles.

Alternate exterior angles, are, essentially, exterior angles of a transversal that runs through two parallel lines. These angles are on alternating sides. These angles, assuming the lines are parallel, must be equal.

There's a proof behind why these angles are equal, but I won't bore you with the specifics as the question doesn't require it.

So - we know that these two angles must be equal to prove that the lines A and B are parallel. Knowing this, we can write an equation:

5x + 20 = 3x + 60

Subtract 20 from both sides:

5x + 20 - 20 = 3x + 60 - 20

5x = 3x + 40

Subtract 3x from both sides:

5x - 3x = 3x - 3x + 40

2x = 40

x = 20

If you have any questions on how I got to the answer, just ask!

- breezyツ

3 0

3 years ago

(A)using geometry vocabulary, describe a sequence of transformations that maps figure P (-1,2)(-1,4) (-4,2) (-4,4) onto figure Q

andrey2020 [161]

Before we proceed on determining the transformation happening on this problem, it's better to see first the location of the figure by drawing it in a cartesian coordinate plane. We have

If we observe the figures and the coordinates of the plot, we can see that there is a difference of 1 on the x coordinates of P and y coordinates of Q. Therefore, the first transformation that we consider here is the movement of figure P by 1 unit to the left. We have

$\begin{gathered} P_1=(-1-1,2_{})=(-2,2) \\ P_2=(-1-1,4)=(-2,4) \\ P_3=(-4-1,2)=(-5,2) \\ P_4=(-4-1,4)=(-5,4) \end{gathered}$

This transformation changes the location of figure P into

The next transformation will be the rotation of the red dotted figure on the figure above by 90 degrees counterclockwise. With this transformation, the coordinates will transform as

$P_{ccw,90}=(-y,x)$

Hence, for the rotation, we have the new coordinates.

$\begin{gathered} P_1^{\prime}=(-2,-2) \\ P_2^{\prime}=(-4,-2) \\ P_3^{\prime}=(-2,-5) \\ P_4^{\prime}=(-4,-5) \end{gathered}$

The transformed image, which is represented as NMPO, will now be at

For the last transformation, we will be reflecting the figure NMPO over the y axis. This changes the coordinates as

$P_{\text{rotation,y}-\text{axis}}=(-x,y)$

We now have the new coordinates:

$\begin{gathered} P^{\doubleprime}_1=(2,-2)=Q_1_{}_{} \\ P_2^{\doubleprime}=(4,-2)=Q_3 \\ P_3^{\doubleprime}=(2,-5)=Q_2 \\ P_4^{\doubleprime}_{}=(4,-5)=Q_4_{} \end{gathered}$

As you can see, they have the same coordinates as figure Q.

The mapping rules for the sequence described above are as follows:

First transformation (moving one unit to the left (x-1,y))

$\begin{gathered} P_1(-1,2)\rightarrow P_1(-1-1,2)\rightarrow P_1(-2,2) \\ P_2(-1,4)\rightarrow P_1(-1-1,4)\rightarrow P_2(-2,4) \\ P_3(-4,2)\rightarrow P_1(-4-1,2)\rightarrow P_3(-5,2) \\ P_4(-4,4)\rightarrow P_1(-4-1,4)\rightarrow P_4(-5,4) \end{gathered}$

Second transformation (rotation counter clockwise (-y,x))

$\begin{gathered} P_1(-2,2)\rightarrow P^{\prime}_1(-2,-2)_{} \\ P_2(-2,4)\rightarrow P^{\prime}_2(-4,-2) \\ P_3(-5,2)\rightarrow P^{\prime}_3(-2,-5)_{} \\ P_4(-5,4)\rightarrow P^{\prime}_4(-4,-5)_{} \end{gathered}$

Third Transformation (reflection over y-axis (-x,y))

$\begin{gathered} P^{\prime}_1(-2,-2)\rightarrow P^{\doubleprime}_1(-(-2),-2)\rightarrow P^{\doubleprime}_1=(2,-2)=Q_1 \\ P^{\prime}_2(-4,-2)\rightarrow P^{\doubleprime}_1(-(-4),-2)\rightarrow P^{\doubleprime}_1=(4,-2)=Q_3 \\ P^{\prime}_3(-2,-5)\rightarrow P^{\doubleprime}_1(-(-2),-5)\rightarrow P^{\doubleprime}_1=(2,-5)=Q_2 \\ P^{\prime}_4(-4,-5)\rightarrow P^{\doubleprime}_1(-(-4),-5)\rightarrow P^{\doubleprime}_1=(4,-5)=Q_4 \end{gathered}$

7 0

1 year ago