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2 years ago

15

sandra has 4 math test this marking period. she recieved 90%, 87%. and 92% on the first three. what is the minimum score she nee

ds on her last test in order to average atleast a 90% in math class?

1 answer:

Butoxors [25]2 years ago

6 0

She would need to get a 90% or more.

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Please answer this question

Ann [662]

9.9=9×9=81 and you have to multiply

6 0

3 years ago

The point-Slope form of a line that has a slope of , and passes through the point (3, 0) is shown.

stellarik [79]

Answer:

y=1/4x-3/4

Step-by-step explanation:

You just multiply 4/3 by -3

8 0

3 years ago

Can someone tell me where to put both of the answer questions

Monica [59]

# of tables- y

# of months-x

4 0

2 years ago

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Based on the given angle measures, which triangle has side length measures that could be correct?

Likurg_2 [28]

Answer

The triangle in the figure 4 is correct .

Reason

by using the trignometric identity

$tan\theta = \frac{Perpendicular}{Base}$

Thus

$tan 30^{\circ} = \frac{Perpendicular}{Base}$

$tan 30 ^{\circ} =\frac{1}{\sqrt{3}}$

Now in the figure (1)

$tan 30^{\circ} = \frac{13.9}{8}$

$\frac{1}{\sqrt{3}} = \frac{13.9}{8}$

on simplify

$0.5774 \neq 1.7375$

thus side length measures in the figure (1) is not correct .

Now in the figure (2)

$tan30^{\circ} =\frac{16}{8} \\ \frac{1}{\sqrt{3}} =\frac{16}{8} \\ 0.577 \neq 2$

thus side length measures in the figure (2) is not correct .

Now in the figure (3)

$tan 30 ^{\circ} = \frac{8}{16} \\ \frac{1}{\sqrt{3}} =\frac{8}{16} \\ 0.577\neq0.5$

thus side length measures in the figure (3) is not correct .

Now in the figure (4)

$tan30^{\circ} = \frac{8}{13.9} \\ \frac{1}{\sqrt{3}} =\frac{8}{13.9}\\ 0.577 = 0.576(approx)$

Therefore the figure (4) is correct triangle has side length measures that could be correct

Hence proved

8 0

2 years ago

Read 2 more answers

The first step for deriving the quadratic formula from the quadratic equation, 0 = ax2 bx c, is shown.step 1: –c = ax2 bxwhich b

Ksju [112]

Answer:

option (b) is correct.

First step to solve for deriving the quadratic formula from the quadratic equation is completing the square.

Step-by-step explanation:

Given : the quadratic equation $ax^2+bx+c=0$

We have to find the first step for deriving the quadratic formula from the quadratic equation.

Quadratic formula is given by $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Consider the given quadratic equation $ax^2+bx+c=0$

Divide the equation by a, we get,

$x^2+\frac{b}{a}x+\frac{c}{a}=0$

Take $\frac{c}{a}$ to other side, we get,

$x^2+\frac{b}{a}x=-\frac{c}{a}$

Add $(\frac{b}{2a})^2$ to both side, we get,

$x^2+\frac{b}{a}x+(\frac{b}{2a})^2=(\frac{b}{2a})^2-\frac{c}{a}$

Thus, the left side is in the form of $(a+b)^2=a^2+b^2+2ab$

On solving, we get,

$(x+\frac{b}{2a})^2=(\frac{b}{2a})^2-\frac{c}{a}$

Then solve for x , we get quadratic formula $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Thus, first step to solve for deriving the quadratic formula from the quadratic equation is completing the square.

Thus, option (b) is correct.

3 0

2 years ago

Read 2 more answers