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dsp73
2 years ago
14

Yesterday, Jake drove 43 1/2 miles. Jack used 1 1/4 gallons of gasoline. What is the unit rate for miles per​ gallon?

Mathematics
1 answer:
Thepotemich [5.8K]2 years ago
7 0

Step-by-step explanation:

43.5 miles per 1.25 gallons

= 34.8 miles per gallon.

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Pentagon RSTUV is circumscribed about a circle. What is the value of x if RS = 10, ST = 13, TU = 11, UV = 12, and VR = 12?
Kobotan [32]

Answer:

Step-by-step explanation:

The value of x is 7 ⇒ 1st answer

Step-by-step explanation:

* Lets revise a fact in the circle

- The two tangents drawn from a point out side the circle are equal

∵ RSTUV is circumscribed about a circle

∴ Each side of the pentagon is a tangent to the circle

- Look to the attached figure to know how we will solve the problem

- Each tangent divided into two parts

# RS = x + y

∵ RS = 8

∴ x + y = 8 ⇒ (1)

# RV = x + n

∵ RV = 12

∴ x + n = 12 ⇒ (2)

- Subtract (2) from (1)

∴ y - n = -4 ⇒ (3)

# ST = y + z

∵ ST = 12

∴ y + z = 12 ⇒ (4)

# TU = z + m

∵ TU = 15

∴ z + m = 15 ⇒ (5)

- Subtract (5) from (4)

∴ y - m = -3 ⇒ (6)

# UV = m + n

∵ UV = 9

∴ m + n = 9 ⇒ (7)

- Add (6) and (7)

∴ y + n = 6 ⇒ (8)

- Lets solve equation (3) and equation (8) to find y

∵ y - n = -4 ⇒ (3)

∵ y + n = 6 ⇒ (8)

- Add (3) and (8)

∴ 2y = 2 ⇒ divide two sises by 2

∴ y = 1

- Lets substitute the value of y in equation (1)

∵ x + y = 8 ⇒ (1)

∵ y = 1

∴ x + 1 = 8 ⇒ subtract (1) from both sides

∴ x = 7

* The value of x is 7

plz mark me as brainlyist thxx

7 0
2 years ago
Read 2 more answers
Ac=4 bx=y+4 what is the value of y
Kitty [74]
Ac=4 bx=y+4 doesn't make sense.  Perhaps you meant <span>Ac-4bx=y+4.  If this is not correct, ensure that you have copied down the original problem correctly.

</span>To solve  Ac-4bx=y+4  for y, subtract 4 from both sides of this equation.  You'll get:

Ac-4bx-4 = y+4 - 4, or y = ac-4bx - 4.
6 0
3 years ago
Help me on this please
zalisa [80]

Answer:

1. (x, y) → (x + 3, y - 2)

Vertices of the image

a) (-2, - 3)

b) (-2, 3)

c) (2, 2)

2. (x, y) → (x - 3, y + 5)

Vertices of the image

a) (-3, 2)

b) (0, 2)

c) (0, 4)

d) (2, 4)

3. (x, y) → (x + 4, y)

Vertices of the image

a) (-1, -2)

b) (1, -2)

c) (3, -2)

4. (x, y) → (x + 6, y + 1)

Vertices of the image

a) (1, -1)

b) (1, -2)

c) (2, -2)

d) (2, -4)

e) (3, -1)

f) (3, -3)

g) (4, -3)

h) (1, -4)

5. (x, y) → (x, y - 4)

Vertices of the image

a) (0, -2)

b) (0, -3)

c) (2, -2)

d) (2, -4)

6. (x, y) → (x - 1, y + 4)

Vertices of the image

a) (-5, 3)

b) (-5, -1)

c) (-3, 0)

d) (-3, -1)

Explanation:

To identify each <u><em>IMAGE</em></u> you should perform the following steps:

  • List the vertex points of the preimage (the original figure) as ordered pairs.
  • Apply the transformation rule to every point of the preimage
  • List the image of each vertex after applying each transformation, also as ordered pairs.

<u>1. (x, y) → (x + 3, y - 2)</u>

The rule means that every point of the preimage is translated three units to the right and 2 units down.

Vertices of the preimage      Vertices of the image

a) (-5,2)                                   (-5 + 3, -1 - 2) = (-2, - 3)

b) (-5, 5)                                  (-5 + 3, 5 - 2) = (-2, 3)

c) (-1, 4)                                   (-1 + 3, 4 - 2) = (2, 2)

<u>2. (x,y) → (x - 3, y + 5)</u>

The rule means that every point of the preimage is translated three units to the left and five units down.

Vertices of the preimage      Vertices of the image

a) (0, -3)                                   (0 - 3, -3 + 5) = (-3, 2)

b) (3, -3)                                   (3 - 3, -3  + 5) = (0, 2)

c) (3, -1)                                    (3 - 3, -1 + 5) = (0, 4)

d) (5, -1)                                    (5 - 3, -1 + 5) = (2, 4)

<u>3. (x, y) → (x + 4, y)</u>

The rule represents a translation 4 units to the right.

Vertices of the preimage   Vertices of the image

a) (-5, -2)                               (-5 + 4, -2) = (-1, -2)

b) (-3, -5)                               (-3 + 4, -2) = (1, -2)

c) (-1, -2)                                (-1 + 4, -2) = (3, -2)

<u>4. (x, y) → (x + 6, y + 1)</u>

Vertices of the preimage      Vertices of the image

a) (-5, -2)                                  (-5 + 6, -2 + 1) = (1, -1)

b) (-5, -3)                                  (-5 + 6, -3 + 1) = (1, -2)

c) (-4, -3)                                   (-4 + 6, -3 + 1) = (2, -2)

d) (-4, -5)                                  (-4 + 6, -5 + 1) = (2, -4)

e) (-3, -2)                                  (-3 + 6, -2 + 1) = (3, -1)

f) (-3, -4)                                   (-3 + 6, -4 + 1) = (3, -3)

g) (-2, -4)                                  (-2 + 6, -4 + 1) = (4, -3)

h) (-2, -5)                                  (-2 + 3, -5 + 1) = (1, -4)

<u>5. (x, y) → (x, y - 4)</u>

This is a translation four units down

Vertices of the preimage      Vertices of the image

a) (0, 2)                                    (0, 2 - 4) = (0, -2)

b) (0,1)                                      (0, 1 - 4) = (0, -3)

c) (2, 2)                                     (2, 2 - 4) = (2, -2)

d) (2,0)                                     (2, 0 - 4) = (2, -4)

<u>6. (x, y) → (x - 1, y + 4)</u>

This is a translation one unit to the left and four units up.

Vertices of the pre-image     Vertices of the image

a) (-4, -1)                                   (-4 - 1, -1 + 4) = (-5, 3)

b) (-4 - 5)                                  (-4 - 1, -5 + 4) = (-5, -1)

c) (-2, -4)                                  (- 2 - 1, -4 + 4) = (-3, 0)

d) (-2, -5)                                 (-2 - 1, -5 + 4) = (-3, -1)

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