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3 years ago

M = - 3 M = -1 M = 1 M = 3 I’m giving out brainlest to whoever helps me

Mathematics

1 answer:

ASHA 777 [7]3 years ago

4 0

Answer:

I would say 3.

Step-by-step explanation:

I would say 3 because if you look at the graph, it goes: 0, 1, 2, 4, 5. Your missing 3. So, I would say 3.

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In the inequality 1/ 2 x - 6 > 10, x represents Todd's age. Which phrase most accurately describes Todd's age? A) Todd is old

Tom [10]

Answer:

Option A) Todd is older than 32

Step-by-step explanation:

we have

$\frac{1}{2}x-6>10$

Solve for x

Adds 6 both sides

$\frac{1}{2}x>10+6\\\\\frac{1}{2}x>16$

Multiply by 2 both sides

$x>32$

Todd's age is greater than 32

therefore

Todd is older than 32

8 0

2 years ago

What is the slope of a line that is perpendicular to the

alexgriva [62]

Answer:

The slope of a line that is perpendicular to the line

shown in the graph is = 4

Hence, option 'd' is true.

Step-by-step explanation:

From the line equation, let us take two points

(0, 2)

(4, 1)

Finding the slope between two points

$\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(0,\:2\right),\:\left(x_2,\:y_2\right)=\left(4,\:1\right)$

$m=\frac{1-2}{4-0}$

$m=-\frac{1}{4}$

As we know that the slope of the perpendicular line is basically the negative reciprocal of the slope of the line, so

The slope of the perpendicular line will be:

$-\frac{1}{-\frac{1}{4}}=4$

Thus, the slope of a line that is perpendicular to the line

shown in the graph is = 4

Hence, option 'd' is true.

5 0

3 years ago

Which is a quadratic function having a leading coefficient of 3 and a constant term of –12? f(x) = –12x2 3x 1 f(x) = 3x2 11x – 1

eimsori [14]

The correct option is b.

<h2>Quadratic Equation</h2>

A quadratic equation is an equation that can be written in the form of

ax²+bx+c.

Where a is the leading coefficient, and

c is the constant.

<h2>Leading coefficient </h2>

The leading coefficient of an equation is a coefficient of the term having the highest power in the equation.

Given options:

a.) $f(x) = -12x^2+3x+1 \\$

b.) $f(x) = 3x^2+11x-12 \\$

c.) $f(x) = 12x^2+3x+3\\$

d.) $f(x) = 3x -12.$

<h3 /><h3>Solution</h3>

As it is given that the leading coefficient of the equation is 3. therefore, the value of the 'a' in the general quadratic equation is 3,

3x²+bx+c,

Also, the value of the constant is -12, therefore, the value of c is -12,

3x²+bx-12

From all the given options the only feasible option is b only.

Hence, the correct option is b.

Learn more about Quadratic Equation:

brainly.com/question/2263981

8 0

2 years ago

Solve the inequality –1 > –2(x – 4) – 5(4x – 7).

Ksivusya [100]

Answer:

$x > 2$

Step-by-step explanation:

$–1 > –2(x – 4) – 5(4x – 7)$

$–1 > –2x + 8 –20x + 35$

$–1 > - 22x + 43$

$- 1 - 43 > - 22x$

$- 44 > - 22x$

$\frac{ - 44}{ - 22} < \frac{ - 22x}{ - 22}$

$2 < x$

$x > 2$

5 0

3 years ago

23 degrees 20’48 is the same as _degrees.

allochka39001 [22]

Answer:

$\large\boxed{23^o20'48''=\left(23\dfrac{26}{75}\right)^o}$

Step-by-step explanation:

$\text{We know:}\\\\1^o=60'\to1'=\left(\dfrac{1}{60}\right)^o\\\\1'=60''\to1^o=(60)(60'')=3600''\to1''=\left(\dfrac{1}{3600}\right)^o\\\\23^o20'48''\\\\20'=\left(\dfrac{20}{60}\right)^o=\left(\dfrac{1}{3}\right)^o\\\\48''=\left(\dfrac{48}{3600}\right)^o=\left(\dfrac{1}{75}\right)^o\\\\\text{Therefore}\\\\23^o20'48''=23^o+20'+48''=23^o+\left(\dfrac{1}{3}\right)^o+\left(\dfrac{1}{75}\right)^o\\\\=\left(23+\dfrac{1\cdot25}{3\cdot75}+\dfrac{1}{75}\right)^o=\left(23+\dfrac{25}{75}+\dfrac{1}{75}\right)^o=\left(23\dfrac{26}{75}\right)^o$

4 0

3 years ago