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3 years ago

14

Simplify the expression by combining like terms. -3(6j+1)-(3j+6)-4j+7

1 answer:

Alexandra [31]3 years ago

8 0

-18j-1-3j-6-4j+7
-18j-3j-4j-1-6+7
-25j

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A grocery store sells a bag of 3 oranges for $2.01. What is the cost of oranges, in dollars per orange?

Alik [6]

Answer:

0.67

Step-by-step explanation:

2.01 is the total for three oranges.

the math would be 2.01 ÷ 3 = 0.67

to check your answer:

if one orange equals 0.67 and you have three of them

the math would be 3 × 0.67 = 2.01

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2 years ago

Sarah works at a zoo.

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Answer:

An elephant pregnancy lasts around 22 months, meaning that new born baby elephants are not small! A baby elephant is called a calf and can weigh between 200 and 300 lbs and stand about 3 feet (1 m) tall.

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4 years ago

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60 points answer asap

kow [346]

The point slope form of a line is written as y - y1 = m(x -x1)

Where m is the slope and x1 and y1 are the points on the line.

You are told the slope is 3 and the point is (2,-1/2)

x1 is 2 and y1 is -1/2

Replace those in the equation to get:

y- (-1/2) = 3(x-2)

Simplify to get the final answer:

y + 1/2 = 3(x-2)

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3 years ago

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Solve for q. -5(3-q) +4=5q-11

soldier1979 [14.2K]

Perform the indicated multiplication first: -15 + 5q + 4 = 5q - 11

Note that 5q appears on both sides of this equation. Cancelling, we get :

-11 = -11. This is always true. Thus, -5(3-q) +4=5q-11 is true for all q.

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3 years ago

X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?

cluponka [151]

Answer:

$x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

Step-by-step explanation:

I'm assuming the system is $\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.$ :

$x^2+y^2=2$

$x^2+(2x^2-3)^2=2$

$x^2+(4x^4-12x^2+9)=2$

$x^2+4x^4-12x^2+9=2$

$4x^4-11x^2+9=2$

$4x^4-11x^2+7=0$

$x^4-11x^2+28=0$

$(x^2-7)(x^2-4)=0$

$(4x^2-7)(x^2-1)=0$

$4x^2-7=0$

$4x^2=7$

$x^2=\frac{7}{4}$

$x=\pm\sqrt{\frac{7}{4}}$

$x^2-1=0$

$x^2=1$

$x=\pm1$

$y=2x^2-3$

$y=2(\pm\sqrt{\frac{7}{4}})^2-3$

$y=2({\frac{7}{4}})-3$

$y=\frac{7}{2}-3$

$y=\frac{1}{2}$

$y=2x^2-3$

$y=2(\pm1)^2-3$

$y=2(1)-3$

$y=2-3$

$y=-1$

Therefore, $x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

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2 years ago