All categories

3 years ago

The answer to find QS

Mathematics

2 answers:

weqwewe [10]3 years ago

7 0

Since R is the midmoint of QS that means that QR and RS are equal to each other. So first you need to find x. There are many different ways to set up the equation to solve for x, but I'll show you the easiest one:

QR+QS=RT+QR

(2x-4)+(2x-4)=(8x-43)+(2x-4)

Combine like terms:

4x-8=10x-47

Isolate Variable:

4x-8=10x-47

-4x -4x

6x-47=-8

Isolate Constant:

6x-47=-8

+47 +47

6x=39

Divide:

x=6.5

Check Your Answer:

(Plug in 6.5 as x)

(4(6.5)-8)=(8(6.5)-43)+(2(6.5)-4)

18=18

olganol [36]3 years ago

3 0

Answer:

The measure of QS is 68 units.

Step-by-step explanation:

Given information: R is midpoint of QS, RS=2x-4, ST=4x-1, and RT=8x-43.

We need to find the value of QS.

From the given figure it is clear that points Q, R, S and T are collinear.

$RT=RS+ST$

$8x-43=(2x-4)+(4x-1)$

On combining like terms.

$8x-43=(2x+4x)+(-4-1)$

$8x-43=6x-5$

Isolate variable terms.

$8x-6x=43-5$

$2x=38$

Divide both sides by 2.

$x=19$

The value of x is 19.

It is given that R is midpoint of QS. It means R divides the line QS in two equal parts.

$QS=2(RS)$

$QS=2(2x-4)$

Substitute x=19 in the above equation.

$QS=2(2(19)-4)$

$QS=2(38-4)$

$QS=2(34)$

$QS=68$

Therefore the measure of QS is 68 units.

You might be interested in

1/5x-8=4. what does x. equal

Slav-nsk [51]

Answer:

Step-by-step explanation:

if you multiply everything by 5, then you can get rid of the fraction and the equation would now look like this:

x - 40 = 20

and now you can answer this simply

x - 40 = 20

+40 +40

x = 60

and since you multiplied 5 beforehand, now you have to divide by 5 to get the real answer

60 ÷ 5 = 12

therefore, x = 12

8 0

3 years ago

F(3) = 8; f^ prime prime (3)=-4; g(3)=2,g^ prime (3)=-6 , find F(3) if F(x) = root(4, f(x) * g(x))

Marrrta [24]

Given:

$f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6$

Required:

$We\text{ need to find }F^{\prime}(3)\text{ if }F(x)=\sqrt[4]{f(x)g(x)}.$

Explanation:

Given equation is

$F(x)=\sqrt[4]{f(x)g(x)}.$ $F(x)=(f(x)g(x))^{\frac{1}{4}}$ $F(x)=f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}}$

Differentiate the given equation for x.

$Use\text{ }(uv)^{\prime}=uv^{\prime}+vu^{\prime}.\text{ Here u=}\sqrt[4]{f(x)}\text{ and v=}\sqrt[4]{g(x)}.$

$F^{\prime}(x)=f(x)^{\frac{1}{4}}(\frac{1}{4}g(x)^{\frac{1}{4}-1})g^{\prime}(x)+g(x)^{\frac{1}{4}}(\frac{1}{4}f(x)^{\frac{1}{4}-1})f^{\prime}(x)$ $=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}-\frac{1\times4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1}{1}-\frac{1\times4}{4}}f^{\prime}(x)$ $=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1-4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1-4}{4}}f^{\prime}(x)$ $F^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{-3}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{-3}{4}}f^{\prime}(x)$

Replace x=3 in the equation.

$F^{\prime}(3)=\frac{1}{4}f(3)^{\frac{1}{4}}g(3)^{\frac{-3}{4}}g^{\prime}(3)+\frac{1}{4}g(3)^{\frac{1}{4}}f(3)^{\frac{-3}{4}}f^{\prime}(3)$ $Substitute\text{ }f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6\text{ in the equation.}$ $F^{\prime}(3)=\frac{1}{4}(8)^{\frac{1}{4}}(2)^{\frac{-3}{4}}(-6)+\frac{1}{4}(2)^{\frac{1}{4}}(8)^{\frac{-3}{4}}(-4)$ $F^{\prime}(3)=\frac{-6}{4}(8)^{\frac{1}{4}}(2^3)^{\frac{-1}{4}}+\frac{-4}{4}(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}$ $F^{\prime}(3)=\frac{-3}{2}(8)^{\frac{1}{4}}(8)^{\frac{-1}{4}}-(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}$ $F^{\prime}(3)=\frac{-3}{2}\frac{\sqrt[4]{8}}{\sqrt[4]{8}}-\frac{\sqrt[4]{2}}{\sqrt[4]{8^3}}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^9}}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^4(2)^4}(2)}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{4\sqrt[4]{}(2)}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{1}{4}$ $F^{\prime}(3)=\frac{-3\times2}{2\times2}-\frac{1}{4}$ $F^{\prime}(3)=\frac{-6-1}{4}$ $F^{\prime}(3)=\frac{-7}{4}$

Final answer:

$F^{\prime}(3)=\frac{-7}{4}$

8 0

1 year ago

joseph and patrick purchase school supplies in the book store. Joseph purchases four notebooks and three pens for $10.65. Patric

natali 33 [55]

I think it is going to be 3.55 for each notebook and 1.9 for each pen

6 0

3 years ago

What are the value(s) for K? 7K^3=56 Please show your work

Phoenix [80]

I hope this helps you