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3 years ago

Help pls I need help with this lol

Mathematics

1 answer:

Nezavi [6.7K]3 years ago

5 0

Answer:

B. 70

Step-by-step explanation:

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Find the missing side lengths. All answers should be in fully simplified form. a=? b?

Alekssandra [29.7K]

Answer:

a=8 b=6

Step-by-step explanation:

Using the Pythagorean theorem, a^2+b^2=c^2, and in this case, a^2+b^2=10^2 = 100

so a^2+b^2=100

8^2+6^2=100

64+36=100 ✔

and since A is longer than B, a=8, b=6

7 0

3 years ago

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7n + 5 =68.... how to solve this problem and show my work

mart [117]

Answer:

n=9

Step-by-step explanation:

7n+5=68 Subtract 5 on both sides.

-5 -5

7n=63 Divide both sides by 7.

/7 /7

n=9

Hope this helps! Have a nice day ^^

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3 years ago

Read 2 more answers

Complete the table for the given rule. Rule: y = 6x-4 Please help

d1i1m1o1n [39]

y = 6x - 4

Substitute the given x values to solve for y.

x = 1:

y = 6(1) - 4

y = 6 - 4

y = 2

1 = 2

x = 3:

y = 6(3) - 4

y = 18 - 4

y = 14

3 = 14

x = 10:

y = 6(10) - 4

y = 60 - 4

y = 56

10 = 56

x y

1 2

3 14

10 56

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20-7x%2B8%5Cleq%205" id="TexFormula1" title="x^{2} -7x+8\leq 5" alt="x^{2} -7x+8\l

nikdorinn [45]

Answer:

x^2-7x+3=0

Step-by-step explanation:

5 0

3 years ago

) Find the coordinates of the point A' which is the symmetric point

soldier1979 [14.2K]

Let, coordinate of point A' is (x,y).

Since, A' is the symmetric point A(3, 2) with respect to the line 2x + y - 12 = 0.

So, slope of line containing A and A' will be perpendicular to the line 2x + y - 12 = 0 and also their center lies in the line too.

Now, their center is given by :

$C( \dfrac{x+3}{2}, \dfrac{y+2}{2})$

Also, product of slope will be -1 .( Since, they are parallel )

$\dfrac{y-2}{x-3} \times -2 = -1\\\\2y - 4 = x - 3\\\\2y - x = 1$

x = 2y - 1

So, $C( \dfrac{2y -1 +3}{2}, \dfrac{y+2}{2})\\\\C( \dfrac{2y + 2}{2}, \dfrac{y+2}{2})$

Also, C satisfy given line :

$2\times ( \dfrac{2y + 2}{2}) + \dfrac{y+2}{2} = 12\\\\4y + 4 + y + 2 = 24\\\\5y = 18\\\\y = \dfrac{18}{5}$

Also,

$x = 2\times \dfrac{18}{5 } - 1\\\\x = \dfrac{31}{5}$

Therefore, the symmetric points is $A'(\dfrac{31}{5}, \dfrac{18}{5})$ .

7 0

3 years ago