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3 years ago

Plz help with this!!!

Mathematics

1 answer:

grigory [225]3 years ago

8 0

Answer:

an exact location in space.

Step-by-step explanation:

P.S. please take a better picture because there is a reflection from the light and it makes it hard to read

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Greata had a price of cloth that is 9 yards long. She cuts it into prices that are each 1/3 yard long. How many prices of cloth

KonstantinChe [14]

Given:

Length of cloth = 9 yards

Length of pieces = $\dfrac{1}{3}$ yards each

To find:

The number of pieces of cloths after cutting.

Solution:

Let n be the number of required pieces.

Length of each piece is $\dfrac{1}{3}$ yards. So, the length of n pieces is $\dfrac{1}{3}n$ yards.

The length of the whole cloth is 9 yards. So, $\dfrac{1}{3}n$ yards must be equal to 9 yards.

$\dfrac{1}{3}n=9$

Multiply both sides by 3.

$n=9\times 3$

$n=27$

Therefore, the Greata have 27 pieces of cloths.

5 0

3 years ago

25 POINTS: In 1960, 21.5% of U.S. households did not have a telephone. This statistic decreased by 75.8% between 1960 and 1990.

dem82 [27]

Answer: 94.8 % (approx)

Step-by-step explanation:

Let the total household in the U.S is x,

⇒ The total household who did not have a telephone in 1960 = 21.5 % of x

= 0.215 x

According to the question,

The total household who do not have a telephone in 1990 = 75.8 % less than the household in 1960 who did not have the cell phone,

= (100-75.8)% of 0.215x

= 24.2 % of 0.215x

= 0.05203x,

⇒ The household in 1990 who had a telephone = x - 0.05203x = 0.94797x

Hence, the percentage of household who has a telephone in 1990

= $\frac{0.94797x}{x}\times 100$

= $94.797\%\approx 94.8\%$

⇒ Around 94.8% household had telephone in the US in 1990.

7 0

4 years ago

In 2018, Gretal earned $32 000.

Aleonysh [2.5K]

Answer:

she paid $7860

PLS MARK ME AS BRAINLIAST

AND DON'T FORGET TO FOLLOW AND THANKS MY ANSWER

HOPE IT HELPS.....

5 0

3 years ago

Read 2 more answers

35. Which side of the triangle in the diagram is the hypotenuse?<br> A. X<br> B. Z<br> C. C

sleet_krkn [62]

Answer:

the answer in the image is Y

4 0

3 years ago

Read 2 more answers

Find the solutions to the following linear-quadratic systems algebraically. Select the ordered pair that is one of the correct s

Ghella [55]

Answer:

(-2, 6)

Step-by-step explanation:

we have

$y=x^{2} +3x+8$ ----> equation A

$y=2x+10$ ----> equation B

equate equation A and equation B

$x^{2} +3x+8=2x+10$

$x^{2} +3x+8-2x-10=0$

$x^{2} +x-2=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2} +x-2=0$

$a=1\\b=1\\c=-2$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(1)(-2)}} {2(1)}$

$x=\frac{-1\pm\sqrt{9}} {2}$

$x=\frac{-1\pm3} {2}$

$x=\frac{-1+3} {2}=1$

$x=\frac{-1-3} {2}=-2$

Find the values of y

For x=1

$y=2(1)+10=12$

For x=-2

$y=2(-2)+10=6$

therefore

The solutions are (1,12) and (-2,6)

3 0

3 years ago