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Gelneren [198K]

4 years ago

14

Jasmin's dog weighs 55 pounds. Her vet told her that a healthy weight for her dog would be less than or equal to 42 pounds. Jasm

in's dog can lose an average of 7 pounds every week. Write the solution to an inequality that describes all possible number of weeks x for which her dog can be at a healthy weight.

2 answers:

Mnenie [13.5K]4 years ago

8 0

Answer:

x = 2

Step-by-step explanation:

55-42=13

13 brought up to 14 so it can be divided by 7

14 / 7 = 2

maria [59]4 years ago

8 0

The correct answer x=2
Explanation -14/7= 2

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What is the answer to -7(7m-4)=77

il63 [147K]

<span>-7(7m-4)=77
Use distributive property(Multiply -7 by 7m and -4)
-49m + 28= 77
Subtract 28 from both sides
-49m= 49
Divide -49 on both sides so that the only thing remaining on the left side is the variable m.
Final Answer: m = -1</span>

4 0

3 years ago

The fraction a/b is in simplest terms and equal to the following:((2020^3)-1)/((2021^3)+1) . What is a + b?

blagie [28]

Consider the expression

$\dfrac{x^3-1}{(x+1)^3+1}$

Factorize the numerator and denominator a difference/sum of cubes:

$\dfrac{(x-1)(x^2+x+1)}{((x+1)+1)((x+1)^2-(x+1)+1)}$

Expand the denominator:

$\dfrac{(x-1)(x^2+x+1)}{(x+2)((x^2+2x+1)-(x+1)+1)}=\dfrac{(x-1)(x^2+x+1)}{(x+2)(x^2+x+1)}=\dfrac{x-1}{x+2}$

since <em>x</em> = 2020, and clearly 2020² + 2020 + 1 ≠ 0, so we can cancel the factor of <em>x</em> ² + <em>x</em> + 1. This leaves us with

$\dfrac{2020^3-1}{2021^3+1} = \dfrac{2020-1}{2020+2} = \dfrac{2019}{2022}=\dfrac{673}{674}=\dfrac ab$

so that <em>a</em> + <em>b</em> = 673 + 674 = 1347.

7 0

3 years ago

Problems 2.21, 2.22, 2.23

RoseWind [281]

Statements can be proved by contrapositive, contradiction or by induction.

<em>2.21 and 2.23 are proved by contrapositive</em>
<em>2.22 is proved by induction</em>

<u />

<u />

<u>2.21: If </u> $n^3$ <u> is even, then n is even (By contrapositive)</u>

The contrapositive of the above statement is that:

<em>If n is odd, then </em> $n^3$ <em> is odd</em>

Represent the value of n as:

$n = 2k + 1$ , where $k \ge 0$

Take the cube of both sides

$n^3 = (2k + 1)^3$

Expand

$n^3 = 8k^3 + 6k^2 + 6k + 1$

Group

$n^3 =[ 8k^3 + 6k^2 + 6k] + 1$

Factor out 2

$n^3 =2[4k^3 + 3k^2 + 3k] + 1$

Assume w is an integer; where:

$w =4k^3 + 3k^2 + 3k$

So, we have:

$n^3 =2w + 1$

The constant term (i.e. 1) means that $n^3$ is odd.

Hence, the statement has been proved by contrapositive.

<em>i.e. If n is odd, then </em> $n^3$ <em> is odd</em>

<u />

<u>2.22 </u> $3n + 4$ <u> is even, if and only if n is even</u>

We have: $3n + 4$ <u />

<u />

Assume that: $n = 2k + 2$ for $k \ge 0$

So, we have:

$3n + 4 = 3(2k + 2) + 4$

Open bracket

$3n + 4 = 6k + 6 + 4$

$3n + 4 = 6k + 10$

Factorize

$3n + 4 = 2(3k + 5)$

The factor of 2 means that $3n + 4$ is even.

<em>Hence, </em> $3n + 4$ <em> is even, if and only if n is even </em>

<em />

<u />

<u>2.22: </u> $s \ne -1$ <u> and </u> $t \ne -1$ <u>, then </u> $s + t + st \ne -1$ <u />

To do this, we prove by contrapositive.

The contrapositive of the above statement is:

If $s = -1$ and $t=-1$ , then $s + t + st = -1$

We have:

$s + t + st = -1$

Substitute the values of s and t in: $s + t + st = -1$

$-1 -1 -1 \times -1 = -1$

$-1 -1 + 1 = -1$

$-1 = -1$

Hence, by contrapositive:

If $s = -1$ and $t=-1$ , then $s + t + st = -1$

Read more about proofs at:

brainly.com/question/19643658

7 0

3 years ago

6 = 2(1+2)<br> How do I do this?????????????????????

slega [8]

Answer:

6=6

Step-by-step explanation:

6=2+4

6=6

8 0

3 years ago

Read 2 more answers

............??? I need answers

jeka94

On #1, x= 62°
Because you would divide 124° by 2

On #2, x=139°
Because since it is the center point it is the same

Hope this helped :)

5 0

4 years ago