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mel-nik [20]
4 years ago
9

Pls help (>_<) im soo stupid :-[

Mathematics
2 answers:
a_sh-v [17]4 years ago
6 0
Assume that sugar 2 X, flour 3 X
Solving: 2 X= 6
                 X= 3
            So 3 X= 3 · 3=9
Answer: There're 9 cups of flours.
Bad White [126]4 years ago
3 0
You set up a ratio:
\frac{2\ sugar}{3\ flour}=\frac{6\ cups}{x\ cups}
Cross-multiply:
\frac{2\ sugar}{3\ flour}=\frac{6\ cups}{x}\\&#10;2*x=3*6\ cups\\&#10;x=\frac{18\ cups}{2}\\&#10;x=9\ cups
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Which equation represents a line parallel to<br> the line y = 3x - 2?
Vera_Pavlovna [14]

Answer:

y=3x+2

Step-by-step explanation:

-2's opposite is positive and the rate of change (slope) is constant

4 0
3 years ago
Which expression is equivalent to 6x+7-12*2-(3 to the power 2 +3)-x
kow [346]

Step-by-step explanation:

Questions about equivalent expressions usually feature both simple expressions and complex expressions. To check which complex expression is equivalent to the simple expression:

Distribute any coefficients: a(bx\pm c)=abx\pm aca(bx±c)=abx±aca, left parenthesis, b, x, plus minus, c, right parenthesis, equals, a, b, x, plus minus, a, c.

Combine any like terms on each side of the equation: xxx-terms with xxx-terms and constants with constants.

Arrange the terms in the same order, usually xxx-term before constants.

If all of the terms in the two expressions are identical, then the two expressions are equivalent.

Example

How do we solve for unknown coefficients?

Some questions will present us with an equation with algebraic expressions on both sides. On one side, there will be an unknown coeffient, and the question will ask us to find its value.

For the equation to be true for all values of the variable, the two expressions on each side of the equation must be equivalent. For example, if ax+b=cx+dax+b=cx+da, x, plus, b, equals, c, x, plus, d for all values of xxx, then:

aaa must equal ccc.

bbb must equal ddd.

To find the value of unknown coefficients:

Distribute any coefficients on each side of the equation.

Combine any like terms on each side of the equation.

Set the coefficients on each side of the equation equal to each other.

Solve for the unknown coefficient.

Example

How do we rearrange formulas?

Formulas are equations that contain 222 or more variables; they describe relationships and help us solve problems in geometry, physics, etc.

Since a formula contains multiple variables, sometimes we're interested in writing a specific variable in terms of the others. For example, the formula for the area, AAA, for a rectangle with length lll and width www is A=lwA=lwA, equals, l, w. It's easy to calculate AAA using the formula if we know lll and www. However, if we know AAA and www and want to calculate lll, the formula that best helps us with that is an equation in which lll is in terms of AAA and www, or l=\dfrac{A}{w}l=

w

A

l, equals, start fraction, A, divided by, w, end fraction.

Just as we can add, subtract, multiply, and divide constants, we can do so with variables. To isolate a specific variable, perform the same operations on both sides of the equation until the variable is isolated. The new equation is equivalent to the original equation.

7 0
3 years ago
Read 2 more answers
The Houston rockets scored 66 points in the second half. There are 24 minutes in a half. What was the average number of points p
anastassius [24]
The answer is 2.75, I believe.
3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Cpi%20%20x%5E%7B2%7D%20" id="TexFormula1" title=" \pi x^{2} " alt=" \pi x^{2} " align="
ANEK [815]

The formule is about area of a circle.

\pi x^{2} = a

We need to solve the equation for x. Where x is the radius of the circle.

In order to solve it for x, we need isolate it for x on left side.

So, first we need to get rid pi from left side.

On dividing both sides by pi, we get

\frac{\pi x^2}\pi}=\frac{a}{\pi}

x^2=\frac{a}{\pi}

Taking square root on both sides, we get

\sqrt{x^2}=\sqrt{\frac{a}{\pi}}

x=\sqrt{\frac{a}{\pi}}


4 0
4 years ago
Simplify 10+4(−8q−4)
vlada-n [284]

Answer:

10 + 4(-8q - 4)

= 10 + 4 * (-8q) + 4 * (-4)     (Distribute 4)

= 10 - 32q - 16                    (Expand)

= -32q - 6                           (Combine like terms)

4 0
4 years ago
Read 2 more answers
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