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3 years ago

Help please! Need help with #6.

Mathematics

2 answers:

umka21 [38]3 years ago

5 0

You had a y amount of dollars and gave 4 away, now you have less than Cindy. Cindy only has 2 dollars.

podryga [215]3 years ago

3 0

Uppsss no se yo también estoy en eso

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Point B is between points A and C. Find the length of AC if AB = 3x - 2, BC = 2x + 7 and AC = 4x + 10

kirill115 [55]

Answer:

AC = 30

Step-by-step explanation:

Start with:

$4x+10=3x-2+2x+7$

Combine like terms.

$4x+10=5x+5$

Subtract $4x$ from both sides of the equation.

$10=x+5$

Subtract $5$ from both sides of the equation.

$x = 5$

Then, substitute $5$ for $x$ in $AC=4x+10$

$AC=4(5)+10$

Multiply.

$AC=20+10$

Finally, add.

$AC=30$

3 0

3 years ago

What is the slope of the line shown in the graph?

Katarina [22]

Answer:

I'm pretty sure it's -3/2

6 0

3 years ago

Read 2 more answers

Find the solution set of the following quadratic equations using the quadratic formula.

oksian1 [2.3K]

Answer:

<h3> 1) $x_1=\dfrac{7+\sqrt{13}}2\,,\quad x_2=\dfrac{7-\sqrt{13}}2$ </h3><h3> 2) $x_1=-\dfrac13\,,\quad x_2=-3$ </h3>

Step-by-step explanation:

$x^2 - 7x + 9 = 0\quad\implies\quad a=1\,,\ b = -7\,,\ c=9\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-7)\pm\sqrt{(-7)^2-4\cdot1\cdot9}}{2\cdot1}=\dfrac{7\pm\sqrt{49-36}}2\\\\x_1=\dfrac{7+\sqrt{13}}2\,,\quad x_2=\dfrac{7-\sqrt{13}}2$

<h3>2)</h3><h3> $3x^2 + 10x=-3\\\\3x^2+10x+3=0\quad\implies\quad a=3\,,\ b =10\,,\ c=3\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-10\pm\sqrt{10^2-4\cdot3\cdot3}}{2\cdot3}= \dfrac{-10\pm\sqrt{100-36}}6\\\\x_1=\dfrac{-10+\sqrt{64}}6=\dfrac{-10+8}6=-\dfrac13\,,\qquad x_2=\dfrac{-10-8}6=-3$ </h3>