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3 years ago

Um help me please !!!

Mathematics

1 answer:

Aloiza [94]3 years ago

3 0

Answer:

x = 15

Step-by-step explanation:

we know that a straight line have 180 angle

mean the sum of all angle is 180 degree

54 + x + 111 = 180

x = 15

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Using prime factorization what is the GCF 3y 2 squared, 24y 3 cubed

WINSTONCH [101]

Answer:

$3y^2$

Step-by-step explanation:

Given:

To find the GCF of $3y^2\ and\ 24y^3$ using prime factorization.

Writing each in prime factors:

3 = 1 $\times$ 3

24 = 1 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 3

Now, GCF of 3 and 24 is 3

$y^2=1\times y\times y$

$y^3=1\times y\times y\times y$

GCF of $y^2$ and $y^3$ is $y\times y=y^2$ .

Therefore, the overall GCF of the two terms is $3y^2$

8 0

3 years ago

Find the output y when the input x is 5 y=5x-3

jasenka [17]

Answer:

y=22

Step-by-step explanation:

y=5x-3

y=5(5)-3

y=25-3

y=22

3 0

3 years ago

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--- What is the slope of the equation y=5/4 x -7/4

taurus [48]

Answer:

5/4

Step-by-step explanation:

The slope intercept form for a linear equation is y = mx + b. The variable's (x) coefficient, which is represented as m, is the slope of any linear equation.

5 0

3 years ago

I need help with 12 13 and 14

nikdorinn [45]

Answer: Lines $\frac{}{BC}$ and $\frac{}{EF}$ are different lengths.

Step-by-step explanation:

The distance formula is $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }$ , and you can use this formula to solve for the lengths of both lines $\frac{}{BC}$ and $\frac{}{EF}$ .

For line $\frac{}{BC}$ , let $x_{1}$ = the x at point B, or 1, and let $x_{2}$ = the x at point C, or 2.

Now, let $y_{1}$ = the y at point B, or 4, and let $y_{2}$ = the y at point C, or -1.

Now, solve the formula to find the length $\frac{}{BC}$ = $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }\\$ .

$\frac{}{BC}$ = $\sqrt{(1-2)^{2} +(4-(-1))^2$

$\frac{}{BC}$ = $\sqrt{(-1)^{2} +(4+1)^2$

$\frac{}{BC}$ = $\sqrt{1 +5^2$

$\frac{}{BC}$ = $\sqrt{(1+25)$

$\frac{}{BC}$ = $\sqrt{26} \\$

Now, for line $\frac{}{EF}$ , let $x_{1}$ = the x at point E, or -4, and let $x_{2}$ = the x at point F, or -1.

Let $y_{1}$ = the y at point E, or -3, and let $y_{2}$ = the y at point F, or 1.

Now, solve the formula to find the length $\frac{}{EF}$ = $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }\\$ .

$\frac{}{EF}$ = $\sqrt{(-4-(-1))^{2} +(-3-1)^2$

$\frac{}{EF}$ = $\sqrt{(-4+1)^{2} +(-4)^2$

$\frac{}{EF}$ = $\sqrt{(-3)^2+16$

$\frac{}{EF}$ = $\sqrt{(9+16)$

$\frac{}{EF}$ = $\sqrt{25}$

$\frac{}{EF}$ = 5

Now, look back at $\frac{}{BC}$ . The two lines have different lengths, so you have now justified the fact that they are not the same.

Questions 13 and 14 would be solved in much the same way- but please let me know if you want me to show the work for those as well!

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3 years ago

Hi, I have a question: Is 1.2 and 1.20 the same?

Neporo4naja [7]

Answer:

yes

Step-by-step explanation:

0 is just a place holder

8 0

3 years ago

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