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3 years ago

on a SOL, a test that had 60 questions. how many questions could've you gotten wrong to have a passing grade?

Mathematics

2 answers:

iren [92.7K]3 years ago

7 0

I think its 45 or 55

Alisiya [41]3 years ago

5 0

20 questions.

You need a above a 66% to pass and 66% of 60 is 40. That means you could've missed 20 questions and still passed.

Hope that helps!

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Last year, a banquet hall charged $30 per person, and 60 people attended

ohaa [14]

So what’s the question?

6 0

3 years ago

Answer in simplest form

solong [7]

Ok, so you know that we can't divide fractions no matter what. The first thing you do is keep the first fraction, then change the divide sign to multiplication ( opposite of division is multi. ) and finallly flip the last fraction to 7/1 .

Here's a video to help you understand that method more ...
https://www.youtube.com/watch?v=uMz4Hause-o

The answer ends up as 6.53333333333

Hope this helps !

5 0

3 years ago

14. In a right triangle, if one acute angle has measure 10x + 6 and the other has measure 2x + 7, find the larger of these two a

lubasha [3.4K]

Answer:

421/6 degrees

Step-by-step explanation:

The acute angles of a right triangle are complementary, so

$10x+6+2x+7=90 \\ \\ 12x+13=90 \\ \\ 12x=77 \\ \\ x=\frac{77}{12}$

So, the acute angles measures 421/6 degrees and 119/6 degrees, and thus the larger angle is 421/6 degrees.

7 0

1 year ago

Harry is trying to solve the equation y = 2x^2 − x − 6 using the quadratic formula. He has made an error in one of the steps bel

Bond [772]

He made a mistake in step #2. It seemed to be a trivial mistake because it involved signs, but it still had a great impact. Since step#2, his solution was already wrong.

Instead of
(-1)²-4(2)(-6) = 1 + 48 = 49

What he did is
(-1)²-4(2)(-6) = -1 + 48 = 47

4 0

3 years ago

In right △ABC, the altitude CH to the hypotenuse AB intersects angle bisector AL in point D. Find the sides of △ABC if AD = 8 cm

tangare [24]

Answer:

$AC=8\sqrt{3}\ cm\\ \\AB=16\sqrt{3}\ cm\\ \\BC=24\ cm$

Step-by-step explanation:

Consider right triangle ADH ( it is right triangle, because CH is the altitude). In this triangle, the hypotenuse AD = 8 cm and the leg DH = 4 cm. If the leg is half of the hypotenuse, then the opposite to this leg angle is equal to 30°.

By the Pythagorean theorem,

$AD^2=AH^2+DH^2\\ \\8^2=AH^2+4^2\\ \\AH^2=64-16=48\\ \\AH=\sqrt{48}=4\sqrt{3}\ cm$

AL is angle A bisector, then angle A is 60°. Use the angle's bisector property:

$\dfrac{CA}{CD}=\dfrac{AH}{HD}\\ \\\dfrac{CA}{CD}=\dfrac{4\sqrt{3}}{4}=\sqrt{3}\Rightarrow CA=\sqrt{3}CD$

Consider right triangle CAH.By the Pythagorean theorem,

$CA^2=CH^2+AH^2\\ \\(\sqrt{3}CD)^2=(CD+4)^2+(4\sqrt{3})^2\\ \\3CD^2=CD^2+8CD+16+48\\ \\2CD^2-8CD-64=0\\ \\CD^2-4CD-32=0\\ \\D=(-4)^2-4\cdot 1\cdot (-32)=16+128=144\\ \\CD_{1,2}=\dfrac{-(-4)\pm\sqrt{144}}{2\cdot 1}=\dfrac{4\pm 12}{2}=-4,\ 8$

The length cannot be negative, so CD=8 cm and

$CA=\sqrt{3}CD=8\sqrt{3}\ cm$

In right triangle ABC, angle B = 90° - 60° = 30°, leg AC is opposite to 30°, and the hypotenuse AB is twice the leg AC. Hence,

$AB=2CA=16\sqrt{3}\ cm$

By the Pythagorean theorem,

$BC^2=AB^2-AC^2\\ \\BC^2=(16\sqrt{3})^2-(8\sqrt{3})^2=256\cdot 3-64\cdot 3=576\\ \\BC=24\ cm$

3 0

2 years ago