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3 years ago

Unit rate of 1/2 hour:$1.25

1 answer:

4 0

What I would do with is just multiply the 125 by half an hour yeah or just multiply the ones we five by in half an hour and get your final answer I multiply the 1.25 by 1.2 and I got 1.5 + If I multiply the 1.25 + 2i would get 2.5

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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 $\sf\:x_{2}$ .

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

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

3 years ago

What is the value of the function when x = 3?<br><br> x −2 −1 1 3 4<br> y 4 3 6 5 8

Anon25 [30]

It tells us that:

(-2,4)
(-1,3)
(1,6)
(3,5)
(4,8)

Look at (3,5) ... it tells us that x = 3 and y = 5

So y = 5

6 0

3 years ago

Share £15 in 2:3 <br> Please help me put

son4ous [18]

2 ratio 3,so, 2+3=5
15/5=3 add and then division

5 0

3 years ago

How does the graph change frorm point G to point K?

rewona [7]

Answer:

The graph increases then decreases

Step-by-step explanation:

If you look at point G it is increasing until it gets to the x value of 0. From there on it is decreasing to K.

3 0

3 years ago

Read 2 more answers

The number of hours (H) that a candle will burn increases when the length of the candle (L) increases. Write the correct equatio

jeka94

$\bf \qquad \qquad \textit{direct proportional variation}\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\
%. Length Hours 15 3 20 4
\begin{array}{ccllll}
length&hours\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
15&3\\
20&4
\end{array}\\\\
-------------------------------\\\\$

$\bf \textit{H increases as L increases}\implies H=kL
\\\\\\
\textit{from the table above, we also know that }
\begin{cases}
L=15\\
H=3
\end{cases}
\\\\\\
3=k15\implies \cfrac{3}{15}=k\implies \cfrac{1}{5}=k\quad thus\quad \boxed{H=\cfrac{1}{5}L}
\\\\\\
\textit{now, what is H when L = 2}\qquad H=\cfrac{1}{5}\implies \cdot 2$

4 0

3 years ago

Read 2 more answers