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3 years ago

Y+7=-2(x-1) hurry plz

Mathematics

1 answer:

zhannawk [14.2K]3 years ago

8 0

Answer:

y=-2x-5

Step-by-step explanation:

First of all, you have to distribute -2 to (x-1) to get to this equation y+7=-2x+2. Then, you subtract 7 on both sides to get the slope-intercept of y = -2x - 5.

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Citrus2011 [14]

Answer:

3* (2 x 8) = 3 * (8 x 2)

Step-by-step explanation:

multiplication distributes over addition.

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3 years ago

Write and solve an equation to find the unknown side length x (in inches). Perimeter =24.2 in. Sides of the shape are 8.3, 8.3,

Anastasy [175]

Answer:

Perimeter of shape,which is a convex polygon i.e quadrilateral = 24.2 in

Sides are 8.3 in, 8.3 in, 3.8 in ,and length of fourth side be x in.

As , Perimeter = Sum of all sides

→ 24.2 = 8.3 + 8.3 + 3.8 + x

→ 24.2 = 16.6 +3.8 + x

→ 24.2 = 20 .4 + x

Keeping like terms on one side, of equation

→ 24.2 - 20.4 = x

→ 3.8 = x

→ x = 3.8

So, length of fourth side = 3.8 in

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3 years ago

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lina2011 [118]

Answer:

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Step-by-step explanation:

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3 years ago

An Xbox that normally sells for $390 is on sale at a 30% discount. What is the sale price of the Xbox?

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3 years ago

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$