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makvit [3.9K]
3 years ago
15

Anyone can help me out on this? :)

Mathematics
1 answer:
Charra [1.4K]3 years ago
6 0

Answer:

the answer is c, science CANNOT answer all questions

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Can someone help me with this
Morgarella [4.7K]

Answer:

FG=30

Step-by-step explanation:

Since we know that Point G is on the Segment FH, it doesn't really matter where G is, but we can know for certain that:

FH=FG+GH

We are given that FH is 4x, GH is x, and FG is 2x+10. Substitute:

4x=(2x+10)+x

Solve for x. On the right, combine like terms:

4x=3x+10

Subtract 3x from both sides:

x=10

So, the value of x is 10.

To find the value of FG, substitute 10 into its x:

FG=2x+10\\FG=2(10)+10

Multiply:

FG=20+10

Add:

FG=30

And we're done!

6 0
3 years ago
Pls solve this sum asap
GalinKa [24]
This is a square root meaning whatever number multiplied by itself equals it’s square root. The answer would be 4 on the outside and 3 on the inside

6 0
3 years ago
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Use the listing method to represent the following set.
Dafna11 [192]
The answer is c bc x is 3 and more
7 0
3 years ago
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1+-w2+9w and I need help cuz I’m on 76 and I’m sooo close help
Gnesinka [82]

\huge \boxed{\mathfrak{Question} \downarrow}

  • Simplify :- 1 + - w² + 9w.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

\large \sf1 + - w ^ { 2 } + 9 w

Quadratic polynomial can be factored using the transformation \sf \: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where \sf x_{1} and x_{2} are the solutions of the quadratic equation \sf \: ax^{2}+bx+c=0.

\large \sf-w^{2}+9w+1=0

All equations of the form \sf\:ax^{2}+bx+c=0 can be solved using the quadratic formula: \sf\frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

\large \sf \: w=\frac{-9±\sqrt{9^{2}-4\left(-1\right)}}{2\left(-1\right)}  \\

Square 9.

\large \sf \: w=\frac{-9±\sqrt{81-4\left(-1\right)}}{2\left(-1\right)}  \\

Multiply -4 times -1.

\large \sf \: w=\frac{-9±\sqrt{81+4}}{2\left(-1\right)}  \\

Add 81 to 4.

\large \sf \: w=\frac{-9±\sqrt{85}}{2\left(-1\right)}  \\

Multiply 2 times -1.

\large \sf \: w=\frac{-9±\sqrt{85}}{-2}  \\

Now solve the equation \sf\:w=\frac{-9±\sqrt{85}}{-2} when ± is plus. Add -9 to \sf\sqrt{85}.

\large \sf \: w=\frac{\sqrt{85}-9}{-2}  \\

Divide -9+ \sf\sqrt{85} by -2.

\large \boxed{ \sf \: w=\frac{9-\sqrt{85}}{2}} \\

Now solve the equation \sf\:w=\frac{-9±\sqrt{85}}{-2} when ± is minus. Subtract \sf\sqrt{85} from -9.

\large \sf \: w=\frac{-\sqrt{85}-9}{-2}  \\

Divide \sf-9-\sqrt{85} by -2.

\large \boxed{ \sf \: w=\frac{\sqrt{85}+9}{2}}  \\

Factor the original expression using \sf\:ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \sf\frac{9-\sqrt{85}}{2}for \sf\:x_{1} and \sf\frac{9+\sqrt{85}}{2} for \sf\:x_{2}.

\large \boxed{ \boxed {\mathfrak{-w^{2}+9w+1=-\left(w-\frac{9-\sqrt{85}}{2}\right)\left(w-\frac{\sqrt{85}+9}{2}\right) }}}

<h3>NOTE :-</h3>

Well, in the picture you inserted it says that it's 8th grade mathematics. So, I'm not sure if you have learned simplification with the help of biquadratic formula. So, if you want the answer simplified only according to like terms then your answer will be ⇨

\large \sf \: 1 + -  w {}^{2}  + 9w \\  =\large  \boxed{\bf \: 1 -  {w}^{2}   + 9w}

This cannot be further simplified as there are no more like terms (you can use the biquadratic formula if you've learned it.)

4 0
2 years ago
What is the mean absolute deviation of the data set
Ludmilka [50]

Answer:

Mean absolute deviation. The mean absolute deviation of a dataset is the average distance between each data point and the mean. It gives us an idea about the variability in a dataset

Step-by-step explanation:

Hope this helps

7 0
2 years ago
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