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3 years ago

Use the distributive property to remove the parentheses (-6+4v+2y)(-2)

Mathematics

2 answers:

natima [27]3 years ago

8 0

Answer: -8v-4y+12

Step-by-step explanation:

liubo4ka [24]3 years ago

3 0

Answer:

-8v-4y+12

Step-by-step explanation:

(-2)(-6+4v+2y)

-2 times -6 equals 12

-2 times +4v equals -8v

-2 times +2y equals -4y

So the answer is -8v-4y+12

Hope this helps!

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Solve for x. x + 1 = 2 Ox= 1/4 O x= 1/2 O x = -1/2 0 x = -1/4

Bogdan [553]

Answer:

x=-1/2

Step-by-step explanation:

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3 years ago

Solve for x. 3^x−1=9^x+2 Enter your answer in the box.

slava [35]

Answer:

-5 =x

Step-by-step explanation:

3^(x-1) = 9^(x+2)

Replace 9 with 3^2

3^(x-1) = 3^2^(x+2)

We know that a power to a power means the powers are multiplied

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3^(x-1) = 3^(2x+2)

When the bases are the same, the powers have to be the same

x-1 = 2x+4

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4 0

3 years ago

LenaIs is converting 2/25 to a percent what should she multiply the numerator and the denominator of the fraction why to find th

Contact [7]

Answer:

Multiply 2/25 by 4 to get 8/100 which can be turned into 8%

Step-by-step explanation:

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2 years ago

Wday is the area in square inches of the floor in the scale drawing

Mice21 [21]

Answer:

Step-by-step explanation:

Since 24 is the original length, you would have to see what number to divide by to get 6.

24 divided by 4 is 6.

Now we know that 1/4 is the scale factor, so apply that to 32.

32 divided by 4 is 6

That's your answer!

Brainliest would v much be appreciated

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3 years ago

Let f (x) = 3x − 1 and ε > 0. Find a δ > 0 such that 0 < ∣x − 5∣ < δ implies ∣f (x) − 14∣ < ε. (Find the largest

s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and $\varepsilon>0$ , we want to find a $\delta$ such that $|f(x)-14|$

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that $|x-5|$ . Then

$|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta$ .

Then, if $|x-5|$ , then $|f(x)-14|$ . So, in this case, if $3\delta \leq \varepsilon$ we get that $|f(x)-14|$ . The maximum value of delta is $\frac{\varepsilon}{3}$ .

By definition, this procedure proves that $\lim_{x\to 5}f(x) = 14$ . Note that f(5)=14, so this proves that f is continous at x=5.

3 0

3 years ago