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3 years ago

How do i even start to solve this problem?

Mathematics

2 answers:

Sphinxa [80]3 years ago

8 0

You have to find a and c and then add them all up

nadezda [96]3 years ago

6 0

First you have to find a and c then add them all up

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Find all the points that are on the x-axis that are a distance of 13 units from the point (12,-5).

Viefleur [7K]

Let point A(x,y) be the point on the x-axis. All points that lie on the x-axis have y-coordinate equal to 0, then A(x,y).

Find the distance from point A to point (12,-5):

$d=\sqrt{(x-12)^2+(0-(-5))^2}=\sqrt{(x-12)^2+25}.$

This distance is equal to 13, then

$\sqrt{(x-12)^2+25}=13,\\ \\(x-12)^2=169-25,\\ \\(x-12)^2=144,\\ \\x-12=12 \text{ or } x-12=-12,\\ \\ x=24 \text{ or }x=0.$

You get two ponts (0,0) and (24,0).

6 0

3 years ago

Mei runs 93.24 miles in 3 weeks. If she runs 6 days each week, what is the average distance she runs each day?

olasank [31]

Answer:

5.18 miles each day

Step-by-step explanation:

3 weeks × 6 days = 18 days

93.24 miles in 18 days gives an average of 93.24 ÷ 18 = 5.18 miles per day

4 0

3 years ago

81, 27, 9 finish the sequence

ololo11 [35]

3, 1, $\frac{1}{3}$ , $\frac{1}{9}$

3 0

4 years ago

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If x= 1/2-√3 , prove that x³-2x²-7x+5=3

prohojiy [21]

$x=\frac{1}{2}-\sqrt3\\\\x^3-2x^2-7x+5=3\\\\(\frac{1}{2}-\sqrt3)^3-2(\frac{1}{2}-\sqrt3)^2-7(\frac{1}{2}-\sqrt3)+5\\\\=(\frac{1}{2})^3-3\cdot(\frac{1}{2})^2\cdto\sqrt3+3\cdot\frac{1}{2}\cdot(\sqrt3)^2-(\sqrt3)^3-...\\\\...-2[(\frac{1}{2})^2-2\cdot\frac{1}{2}\cdot\sqrt3+(\sqrt3)^2]-\frac{7}{2}+7\sqrt3+5$

$=\frac{1}{8}-\frac{3\sqrt3}{4}+\frac{9}{2}-3\sqrt3-2(\frac{1}{4}-\sqrt3+3)-\frac{7}{2}+7\sqrt3+5\\\\=\frac{1}{8}+\frac{9}{2}-\frac{1}{2}-6-\frac{7}{2}+5-\frac{3}{4}\sqrt3-3\sqrt3+2\sqrt3+7\sqrt3\\\\=\frac{1}{8}+\frac{1}{2}-1+5\frac{1}{4}\sqrt3=\frac{1}{8}-\frac{1}{2}+5\frac{1}{4}\sqrt3=\frac{1}{8}-\frac{4}{8}+5\frac{1}{4}\sqrt3\\\\=-\frac{3}{8}+5\frac{1}{4}\sqrt3\neq3$

6 0

3 years ago

Please answer ASAP!!!!

trasher [3.6K]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers