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Alenkinab [10]
3 years ago
14

What’s the x and y intercept? What’s the vertex and axis of symmetry?

Mathematics
1 answer:
Kisachek [45]3 years ago
8 0
Need to see the graph to answer this. There is just a blank coordinate plane.
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What is the value of x? Round to the nearest tenth(1 decimal place)​
liraira [26]

Answer:

x ≈ 15.0 ft

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Equality Properties

<u>Trigonometry</u>

  • tan∅ = opposite over adjacent

Step-by-step explanation:

<u>Step 1: Define</u>

Angle = 28°

Opposite leg of angle = 8 ft

Adjacent leg of angle = x ft

<u>Step 2: Find </u><em><u>x</u></em>

  1. Substitute:                                        tan28° = 8/x
  2. Multiply <em>x </em>on both sides:                 xtan28° = 8
  3. Divide tan28° on both sides:          x = 8/tan28°
  4. Evaluate:                                          x = 15.0458
  5. Round:                                              x ≈ 15.0 ft
6 0
3 years ago
During each hour of exercise, Matthew drinks 1 cups of water. Matthew exercised for 14 hours this
Reil [10]
He drank 420 cups of water in one month
7 0
2 years ago
Put these numbers in order from least to greatest. 0.57 5.07 50.7 57 75 ​
nordsb [41]

Answer:

It's basically already in order: 0.57, 5.07, 50.7, 57, 75

Step-by-step explanation:

Hope it helps!      :3

Plz mark as brainliest!

5 0
3 years ago
Read 2 more answers
Write an equation of the hyperbola given that the center is at (2, -3), the vertices are at (2, 3) and (2, - 9), and the foci ar
zavuch27 [327]
Check the picture below.

so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).

because the hyperbola opens vertically, the fraction with the positive sign will be the one with the "y" in it, like you see it in the picture, so without further adieu,

\bf \textit{hyperbolas, vertical traverse axis }&#10;\\\\&#10;\cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1&#10;\qquad &#10;\begin{cases}&#10;center\ ( h, k)\\&#10;vertices\ ( h,  k\pm a)\\&#10;c=\textit{distance from}\\&#10;\qquad \textit{center to foci}\\&#10;\qquad \sqrt{ a ^2 + b ^2}\\&#10;asymptotes\quad  y= k\pm \cfrac{a}{b}(x- h)&#10;\end{cases}\\\\&#10;-------------------------------

\bf \begin{cases}&#10;h=2\\&#10;k=-3\\&#10;a=6\\&#10;c=2\sqrt{10}&#10;\end{cases}\implies \cfrac{[y- (-3)]^2}{ 6^2}-\cfrac{(x- 2)^2}{ b^2}=1&#10;\\\\\\&#10;\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ b^2}=1&#10;\\\\\\&#10;c^2=a^2+b^2\implies (2\sqrt{10})^2=6^2+b^2\implies 2^2(\sqrt{10})^2=36+b^2&#10;\\\\\\&#10;4(10)=36+b^2\implies 40=36+b^2\implies 4=b^2&#10;\\\\\\&#10;\sqrt{4}=b\implies 2=b\\\\&#10;-------------------------------\\\\&#10;\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 2^2}=1\implies \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 4}=1

3 0
3 years ago
Solve for x in this equation: 3/4+ |5-x| = 13/4
Rom4ik [11]

Answer:

  x = 2.5 or 7.5

Step-by-step explanation:

Subtract 3/4.

  |5 -x| = 10/4 = 2.5

This resolves to two equations:

  • 5 -x = 2.5   ⇒   x = 5 -2.5 = 2.5
  • 5 -x = -2.5   ⇒   x = 5 +2.5 = 7.5

The solutions are x = 2.5 or 7.5.

5 0
3 years ago
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