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3 years ago

REALLY NEED HELP FOR THIS ALGEBRA 2 QUESTION THANK YOU IN ADVANCE

Mathematics

1 answer:

guajiro [1.7K]3 years ago

6 0

Lets make an equation for the line first. The slope is 3 and the y intercept is -1, so we can make the equation y = 3x - 1. Then, to figure out what sign to use (<= or >= due to solid line in graph), we replace the = sign with one and see if a coordinate in the shaded area will furfill the inequality. If it doesn't, we know it needs the other inequality sign. If it does, we have found the correct inequality. So let's try y <= 3x - 1 with the coordinate (1,1). We try it, solve, and get 1 <= 2. So, the inequality for this graph is y <= 3x - 1.

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Someone please help me! Studying for finals and don’t understand how to complete problems like this! Thank you!

Drupady [299]

Answer:

(d). $\frac{152}{377}$

Step-by-step explanation:

sin² β + cos² β = 1

cos (α + β) = cos α cos β - sin α sin β

cos α = √(1 - $\frac{400}{841}$ ) = 21 / 29

cos β = 12 / 13

sin β = √(1 - $\frac{144}{169}$ ) = 5 / 13

sin α = 20 / 29

cos (α + β) = $\frac{21}{29}$ × $\frac{12}{13}$ - $\frac{20}{29}$ × $\frac{5}{13}$ = $\frac{152}{377}$

6 0

3 years ago

Help please and this time please put an explanation just don't only say the answer thank you

coldgirl [10]

✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽

➷ T $\boxed{4}$ B $\boxed{24}$

✽ Ok, so B is the number of boys and T is the number of teachers.

➷ $t \div ( B + 18) = \frac{2}{21}$

$\frac{B}{18} = \frac{4}{3}$

✽ Then solve the second equation for B.

➷ $B = 24$

✽ Plug into the first equation to find T.

➷ $T \div (24 + 18) = \frac{2}{21}$

$T = 4$

✽

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ May ♡

5 0

3 years ago

What the answer and how do I solve it?

tatiyna

Answer:

5xy^2z

Step-by-step explanation:

(-15/-3) x (x^6/x^5) x (y^5/y^3) x z = 5xy^2z

4 0

2 years ago

Read 2 more answers

|x|=7 what is the answer to this equation

wolverine [178]

Answer:

x=7, or -7

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Use the graph of △ABC with midsegments DE, EF and DF. Show that EF is parallel to AC and that EF=1/2 AC

Vinvika [58]

According to the midsegment theorem, the midsegments are parallel to

and half the length of the opposite side.

The completed statement are as follows;

Because the slopes of $\overline{EF}$ and $\overline{AC}$ are both -4, $\overline{EF}$ ║ $\overline{AC}$

EF = $\underline{\sqrt{17}}$ and AC = $\underline{2 \cdot \sqrt{17} }$

Because $\underline{\sqrt{17} }$ = $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17} }$ , $EF = \frac{1}{2} \cdot AC$

Reasons:

From the given graph of ΔABC, we have;

Coordinates of the points A, B, and C are; A(-5, 2), B(1, -2), and C(-3, -6)

The coordinates of the point D and E on $\mathbf{\overline{DE}}$ are; D(-4, -2), and E(-2, 0)

The coordinates of the point F is; F(-1, -4)

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\displaystyle Slope \ of \ line \ \overline{AC} = \mathbf{\frac{(-6) - 2}{-3 - (-5)} = \frac{-8}{2}} = -4$

$\displaystyle Slope \ of \ line \ \overline{EF} = \frac{0 - (-4)}{-2 - (-1)} = \frac{4}{-1} = -4$

$Length \ of \ segment,\ l = \sqrt{\left (x_{2}-x_{1} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}$

Length of EF = √((-1 - (-2))² + (-4 - 0)²) = √(17)

Length of AC = √((-3 - (-5))² + (-6 - 2)²) = √(4 × 17) = 2·√(17)

Therefore, we have;

Because the slope of $\mathbf{\overline {EF}}$ and $\mathbf{\overline {AC}}$ are both , -4, $\overline {EF}$ ║ $\overline {AC}$ . EF = $\underline{\sqrt{17}}$ , and AC

= $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}$ ,. Because $\underline{\sqrt{17}}$ = $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}$ , $EF =\mathbf{ \frac{1}{2} AC}$

Learn more about midsegment theorem of a triangle here:

brainly.com/question/7423948

8 0

2 years ago