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3 years ago

The x-intercept of the line x = 3.5 is . the y-intercept of the line y = -6.5

Mathematics

1 answer:

pshichka [43]3 years ago

4 0

<span>The x-intercept of the line x = 3.5 is (3.5, 0).
The y-intercept of the line y = -6.5 is (0, -6.5)</span>

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morpeh [17]

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3 years ago

I think it’s b but I’m not sure... please show work I need help ASAP

ohaa [14]

The answer is b. Great job! here ill show you the work.
15x +90 > 270
move constant to the right side and change its sign, like this:
15x>270-90
then subtract the numbers
15x>180
lastly, divide both sides by 15.
hope this helped!

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3 years ago

Please answer below :-)

mestny [16]

Answer:

A: 18

Step-by-step explanation:

Hopefully this helps!

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2 years ago

Jill uses some of her honey to make lotion if each bottle of lotion requires 1/4 gallon and she uses a total of 3 gallons how ma

AURORKA [14]

So, we need to count up 1/4 each time till we get to 3.

1/4, 2/4, 3/4, 1, 1 1/4, 1 2/4, 1 3/4, 2, 2 1/4, 2 2/4, 2 3/4, 3.

Now, let's count how many times we counted to get to 3.

12.

So, Jill used 12 bottles of lotion.

Glad I could help, and good luck!

AnonymousGiantsFan

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3 years ago

Read 2 more answers

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

2 years ago