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2 years ago

Someone please help me answer this!!

Mathematics

2 answers:

Butoxors [25]2 years ago

7 0

Answer:

Step-by-step explanation:

im a bit rusty but if you are diving the indices its basically minusing

7-3 = 4

tell me if that doesnt work i have one other way

svet-max [94.6K]2 years ago

5 0

Answer:

x=4

Step-by-step explanation:

7^7/7^3=7^x

7^7-3=7^x

7^4=7^x

2401=7^x

7^4=7^x

4=x

x=4

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The coordinates of the endpoint of QS are Q(-9,8) and S(9,-4). Point R is on cue as such that QR:RS Is in the ratio 1:2. What ar

marishachu [46]

R(–3, 4)

Step-by-step explanation:

Let Q(-9,8) and S(9,-4) be the given points and let R(x, y) divides QS in the ratio 1:2.

By section formula,

$R(x, y)=R\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)$

Here, $x_{1}=-9, y_{1}=8, \text { and } x_{2}=9, y_{2}=-4 \text { and } m=1, n=2$

Substituting this in the section formula

$R(x, y)=R\left(\frac{1(9)+2(-9)}{1+2}, \frac{1(-4)+2(8)}{1+2}\right)$

To simplifying the expression, we get

$\Rightarrow R(x, y)=R\left(\frac{9-18}{3}, \frac{-4+16}{3}\right)$

$\Rightarrow R(x, y)=R\left(\frac{-9}{3}, \frac{12}{3}\right)$

⇒ R(x,y) = R(–3,4)

Hence, the coordinates of point R is (–3, 4).

6 0

3 years ago

Given the formula A = 5h (B + b); solve for B. 2

stellarik [79]

Answer:

B = A/5h - b; You could use

B = (A - 5hb)/5h This just puts everything over a common denominator.

Step-by-step explanation:

A = 5h (B + b) Divide both sides by 5h

A/5h = B + b Subtract b from both sides.

A/5h - b = B

6 0

3 years ago

8 over 15? is it closer to 0, 1 over 2 or 1

ruslelena [56]

8/15 is about 0.53
It would be closer to 1/2 or 1 over 2

4 0

3 years ago

Read 2 more answers

A triangle with sides 11m , 13m and 18m is a right triangle. A True B False

Julli [10]

$\bold{\huge{\pink{\underline{ Solution}}}}$

$\bold{\underline{ Given :-}}$

A triangle with sides 11m, 13m and 18m

$\bold{\underline{ To \: Find :-}}$

We have to check it whether it is right angled triangle or not?

$\bold{\underline{ Let's \: Begin :-}}$

$\sf{\red{ In \:right \:angled\ : triangle, }}$

According to the Pythagoras theorem, The sum of the squares of perpendicular height and the square of the base of the triangle is equal to the square of hypotenuse that is sum of the squares of two small sides equal to the square of longest side of the triangle.

We imply it in the given triangle ,

$\sf{\red{ ( Perpendicular)² + (Base)² = (Hypotenuse)²}}$