Slove the equation -29=9-6k-2k-14

La suma de 5 números consecutivos es 122 cuáles son esos números ?

ra1l [238]

Answer:

1) 22.4

2) 23.4

3) 24.4

4) 25.4

5) 26.4

Step-by-step explanation:

La suma de 5 números consecutivos es 122 ¿cuáles son esos números?

5 números consecutivos se pueden representar como

1) = x

2) x + 1

3) x + 2

4) x + 3

5) x + 4

Por eso:

x + (x + 1) + (x + 2) + (x + 3) + (x + 4) = 122

x + x + 1 + x + 2 + x + 3 + x + 4 = 122

5x +10 = 122

5x = 122 - 10

5 veces = 112

x = 112/5

x = 22.4

5 números consecutivos se pueden representar como

1) x = 22.4

2) x + 1 = 22.4 + 1 = 23.4

3) x + 2 = 22.4 + 2 = 24.4

4) x + 3 = 22.4 + 3 = 25.4

5) x + 4 = 22.4 + 4 = 26.4

7 0

3 years ago

Write a verbal expression for 2(n + 9)

strojnjashka [21]

Answer:

Two times the sum of a number and nine.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Last year Nancy weighted 37 5/8 pounds. This year she weighed 42.7 pounds. How much did she gain?

Andre45 [30]

Answer:

Nancy gained 5.075 pounds.

Step-by-step explanation:

5/8=0.625

37.625

42.7-37.625=5.075

4 0

3 years ago

Read 2 more answers

Angle α lies in quadrant II , and tanα=−125 . Angle β lies in quadrant IV , and cosβ=35 .

Artist 52 [7]

Answer:

$cos(\alpha+\beta)=\frac{33}{65}$

Step-by-step explanation:

step 1

Find cos α

we know that

$tan^2(\alpha)+1=sec^2(\alpha)$

we have

$tan(\alpha)=-\frac{12}{5}$

substitute

$(-\frac{12}{5})^2+1=sec^2(\alpha)$

$sec^2(\alpha)=\frac{144}{25}+1$

$sec^2(\alpha)=\frac{169}{25}$

$sec(\alpha)=\pm\frac{13}{5}$

Remember that Angle α lies in quadrant II

so

sec α is negative

$sec(\alpha)=-\frac{13}{5}$

Find the value of cos α

$cos)\alpha)=\frac{1}{sec(\alpha)}$

so

$cos(\alpha)=-\frac{5}{13}$

step 2

Find sin α

we know that

$tan(\alpha)=\frac{sin(\alpha)}{cos(\alpha)}$

$sin(\alpha)=tan(\alpha)cos(\alpha)$

we have

$tan(\alpha)=-\frac{12}{5}$

$cos(\alpha)=-\frac{5}{13}$

substitute

$sin(\alpha)=(-\frac{12}{5})(-\frac{5}{13})$

$sin(\alpha)=\frac{12}{13}$

step 3

Find sin β

we know that

$sin^2(\beta)+cos^2(\beta)=1$

we have

$cos(\beta)=\frac{3}{5}$

substitute

$sin^2(\beta)+(\frac{3}{5})^2=1$

$sin^2(\beta)=1-(\frac{3}{5})^2$

$sin^2(\beta)=1-\frac{9}{25}$

$sin^2(\beta)=\frac{16}{25}$

$sin(\beta)=\pm\frac{4}{5}$

Remember that

Angle β lies in quadrant IV

so

sin β is negative

$sin(\beta)=-\frac{4}{5}$

step 4

Find cos(α−β)

we know that

$cos(\alpha+\beta)=cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)$

we have

$cos(\alpha)=-\frac{5}{13}$

$cos(\beta)=\frac{3}{5}$

$sin(\alpha)=\frac{12}{13}$

$sin(\beta)=-\frac{4}{5}$

substitute the given values

$cos(\alpha+\beta)=(-\frac{5}{13})(\frac{3}{5})-(\frac{12}{13})(-\frac{4}{5})$

$cos(\alpha+\beta)=(-\frac{15}{65})+(\frac{48}{65})$

$cos(\alpha+\beta)=\frac{33}{65}$

7 0

4 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

gulaghasi [49]

Answer: (C) shifts 6 units to the LEFT

<u>Step-by-step explanation:</u>

The vertex form of an absolute value equation is:

y = a |x - h| + k where;

a is the vertical stretch (irrelevant for this problem)
(h, k) is the vertex

Since h represents the x-coordinate and the x-axis is left to right, then h shifts the graph left or right.

If h is negative, the graph shifts to the left.
If h is positive, the graph shifts to the right.

x + 6 is actually x - (-6), so h is negative and the graph shifts to the left.

8 0

3 years ago