All categories

3 years ago

Help needed with these for problems

Mathematics

1 answer:

Tatiana [17]3 years ago

4 0

I can't read the first one and I also don't know

You might be interested in

Your friend identifies the terms and like terms in the expression

Svetllana [295]

Answer:

The like-terms equation would be:

10x - 5

Step-by-step explanation:

4 0

3 years ago

Can someone please send me an example of this? (Perpendicular)

Sergeu [11.5K]

Ito's like the line BC is vertical and MN is horizontal as we compare both. So BC is perpendicular to MN and MN is also perpendicular To BC

6 0

3 years ago

Read 2 more answers

12 = x - (-9)<br><br><br> What is the variable x

marshall27 [118]

X=3 use photomath it will help

4 0

3 years ago

Find the distance from the point A(-2,3) to the line y=1/2x+1. Round your answer to the nearest tenth.

telo118 [61]

Answer:

Step-by-step explanation:

$\text{Let}\\\\Ax+By=C-\text{a line}\\\\(x_0,\ y_0)-\text{a point}\\\\\text{The formula of a distance between a point and a line}:\\\\d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$

$\text{We have}\\\\y=\dfrac{1}{2}x+1,\ A(-2,\ 3)\\\\\text{Convert the equation of a line to the standard form}:\\\\y=\dfrac{1}{2}x+1\qquad\text{multiply both sides by 2}\\\\2y=x+2\qquad\text{subtract}\ x\ \text{from both sides}\\\\-x+2y=2\qquad\text{change the signs}\\\\x-2y=-2\\\\\text{Substitute}\ A=1,\ B=-2,\ C=-2,\ x_0=-2,\ y_0=3,\ \text{to the formula:}\\\\d=\dfrac{|(1)(-2)+(-2)(3)+(-2)|}{\sqrt{1^2+(-2)^2}}=\dfrac{|-2-6-2|}{\sqrt{1+4}}=\dfrac{|-10|}{\sqrt5}$

$=\dfrac{10}{\sqrt5}\cdot\dfrac{\sqrt5}{\sqrt5}=\dfrac{10\sqrt5}{5}=2\sqrt5$

7 0

3 years ago

Find f(-2) for f(x) = 3 * 2^x

schepotkina [342]

9514 1404 393

Answer:

3/4

Step-by-step explanation:

f(-2) = 3·2^-2 = 3/2^2

f(-2) = 3/4

_____

The applicable rule of exponents is ...

a^-b = 1/a^b

4 0

3 years ago