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2 years ago

PLEASE HELP WITH THIS MATH PROBLEM

Mathematics

1 answer:

Yuki888 [10]2 years ago

4 0

Answer:

-2x<4

Step-by-step explanation:

using the x-values- look for where it starts going up (from left to right) (-2) to the point where it turns back downward (4)

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How do you solve this 13c<-169

Bess [88]

The answer is c < -13

6 0

2 years ago

Find the solution set for this equation.

LuckyWell [14K]

Answer:

m = -5, 0

Step-by-step explanation:

Easiest and quickest way to do this is to graph the equation and find where it intersects the x-axis. If not, then use completing the square or quadratic formula to find your roots.

6 0

3 years ago

Find the missing side of this right a triangle?<br> help pls

yaroslaw [1]

Hi, there!

_____

To find the missing side length of this triangle (which is a <em>right</em> triangle), we'll use Pythagoras' Theorem.

$-\!\!\!-\!\!\!-\!\!\!\longrightarrow\boldsymbol{a^2+b^2=c^2}$

Here,

$\textsc{legs:}\begin{cases} \bold{a}\\\bold{b} \end{cases}$

Hypotenuse (the longest side): c

We know that

$\bold{a=2 \ and \ b=28}$

So, we can substitute our parameters, and solve for c.

$\bold{2^2+28^2=c^2}$ $-\!\!\!-\!\!\!-\!\!\!\longrightarrow$ square 2 and 28

$\bold{4+784=c^2}$ $-\!\!\!-\!\!\!-\!\!\!\longrightarrow$ add the numbers

$\bold{788=c^2}$ $-\!\!\!-\!\!\!-\!\!\!\longrightarrow$ square root both sides

$\bold{\sqrt{788}=c}$

Hope the answer - and explanation - made sense,

happy studying!!

6 0

1 year ago

I need help again.lol

Monica [59]

Answer:

$13\ units$

Step-by-step explanation:

One is asked to find the distance between two points on a coordinate plane. The easiest way to do so is to use the distance formula, this formula is the following:

$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2$

Where ( $(x_1,y_1)$ ) and ( $(x_2,y_2)$ ) are the points which one needs to find the distance between. In this case, these points are as follows:

$k\ (-3,5)$

$j\ (-8, -7)$

Substitute these points into the formula and solve for the distance between these points,

$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2$

$D=\sqrt{((-3)-(-8))^2+((5)-(-7))^2$

Simplify,

$D=\sqrt{((-3)-(-8))^2+((5)-(-7))^2$

$D=\sqrt{(-3+8)^2+(5+7)^2$

$D=\sqrt{(5)^2+(12)^2$

$D=\sqrt{25+144}$

$D=\sqrt{169}$

$D=13$

8 0

2 years ago

Find the area of ABC with vertices A(3,-6), B(5,-6), and C(7,–9).

stellarik [79]

Area of a triangle ABC with the given vertices is 3 square units.

Given that, the vertices of a triangle ABC, A(3,-6), B(5,-6), and C(7,–9).

<h3>What is the area of triangle formula in coordinate geometry?</h3>

In Geometry, a triangle is a three-sided polygon that has three edges and three vertices. The area of the triangle is the space covered by the triangle in a two-dimensional plane.

Area of a triangle = $\frac{1}{2} ( |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|)$

Here, (x1, y1) = A(3,-6), (x2, y2) = B(5,-6), and (x3, y3) = C(7,–9)

Now, the area of a triangle = 1/2 (|3(-6+9)+5(-9+6)+7(-6+6)|)

= 1/2 (|3(3)+5(-3)+7(0)|)

= 1/2 (|(9-15)|)

= 1/2 × 6

= 3 square units

Therefore, area of a triangle ABC with the given vertices is 3 square units.

To learn more about the area of a area of triangle with vertices visit:

brainly.com/question/26633662.

#SPJ1

7 0

1 year ago

PLEASE HELP WITH THIS MATH PROBLEM​

PLEASE HELP WITH THIS MATH PROBLEM