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2 years ago

To all ppl who are good in math, pls answer this I really need ur help...

Mathematics

2 answers:

Alex17521 [72]2 years ago

3 0

Answer:

1 is -4.

4 is a > -44.

3 is x > 1.

2 p is 3

5 is x < 20

6 is -4 < b < 38/7

DENIUS [597]2 years ago

3 0

Answer:

1. m= -4

2. p= 3

3. x is greater than or equal to 1

4. a is greater than or equal to -36

5. x is less than 20

6. b is less than or equal to 38/7 or greater than or equal to -4

Step-by-step explanation:

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Solve |4 - 2x| = |3x+ 2|.

andreyandreev [35.5K]

Answer:

Step-by-step explanation:4 -2x=3x +2

4-2=3x+2x

2=5x

X=2/5

5 0

2 years ago

Read 2 more answers

In the figure below AB is a diameter of circle P. What is the arc measure of minor arc AC in degrees?

arlik [135]

Answer:

$P = 139$

Step-by-step explanation:

See attachment for complete question

Required

Determine measure of Arc length AC

The interpretation of this question is to find P

If AB is the diameter, then

$P + 41 = 180$

$P + 41 - 41= 180 - 41$

$P = 139$

<em>Hence, the measure is 139 degrees</em>

3 0

3 years ago

Please help!

Mrac [35]

Answer:

Equilateral triangles are sometimes

acute triangles.

Scalene triangles are never

acute triangles.

Right triangles are never

acute triangles.

Obtuse triangles are always

isosceles triangles.

3 0

3 years ago

What is the solution to the equation? <br>1/7g=3/14

Crazy boy [7]

Answer:

g = 21/4

Decimal form:

g = 5.25

Mixed Number:

5 1/4

6 0

2 years ago

A ball is thrown from an initial height of 1 meter with an initial upward velocity of 15m/s. The ball's height h (in meters) aft

lakkis [162]

$\bf \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{h}=1+15t-5t^2}\implies \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{6}=1+15t-5t^2}\implies 0=-5+15t-5t^2 \\\\\\ ~~~~~~~~~~~~\textit{using the quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{5}t^2\stackrel{\stackrel{b}{\downarrow }}{-15}t\stackrel{\stackrel{c}{\downarrow }}{+5}=0 \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}$

$\bf t=\cfrac{-(-15)\pm\sqrt{(-15)^2-4(5)(5)}}{2(5)}\implies t=\cfrac{15\pm\sqrt{225-100}}{10} \\\\\\ t=\cfrac{15\pm\sqrt{125}}{10}\implies t=\cfrac{15\pm\sqrt{5^2 \cdot 5}}{10}\implies t=\cfrac{\stackrel{3}{~~\begin{matrix} 15 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\pm ~~\begin{matrix} 5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\sqrt{5}}{\underset{2}{~~\begin{matrix} 10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}$

$\bf t=\cfrac{3\pm \sqrt{5}}{2}\implies t= \begin{cases} \frac{3+ \sqrt{5}}{2} \approx 2.618\\\\ \frac{3- \sqrt{5}}{2}\approx 0.382 \end{cases}$

8 0

2 years ago