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2 years ago

I can't think of any right answer?

Mathematics

2 answers:

madam [21]2 years ago

8 0

I think the answer is true
Hope this helps!!
Please consider brainliest!<3

Inga [223]2 years ago

6 0

It is true

Have a nice day!

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Which lines, if any, can you conclude are parallel given that m<1 + m<2= 8 ? Justify your conclusion with a theorem or pos

Anon25 [30]

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6 0

3 years ago

Sue Mitchell weighed 8 lb. 13 oz. when she was born. She now weighs 93 lb. 5

ra1l [238]

Answer:

Weight gain = 84 lb 5 oz

Average weight gain = 6 lb 5 oz

Step-by-step explanation:

lb means pounds

16 oz = 1 pound

Convert the weight to pounds

Weight when she was born = 8 lb. 13 oz

= 8 pounds + 0.8125 pounds

= 8.8125 pounds

Present weight = 93 lb. 5

= 93 pounds + 0.3125 pounds

= 93.3125 pounds

How much has she gained since birth

Weight gain = Present weight - Weight when she was born

= 93.3125 pounds - 8.8125 pounds

= 84.5 pounds

= 84 lb 5 oz

Weight gain = 84 lb 5 oz

what has been her average weight for each year if she is 13 year old

Average weight gain = Weight gain / 13

= 84.5 pounds / 13

= 6.5 pounds

= 6 lb 5 oz

Average weight gain = 6 lb 5 oz

8 0

2 years ago

Donald bought a box of golf balls for $9.27. There were 18 golf balls in the box.

wlad13 [49]

Answer:

Each golf ball costs about $2.

Step-by-step explanation:

Given:

Donald bought a box of golf balls for $9.27. There were 18 golf balls in the box.

Rename the decimal dividend as a whole number that is compatible with the divisor to estimate the quotient.

Now, to find the cost of each golf ball.

As given:

Rename the decimal dividend as a whole number that is compatible with the divisor to estimate the quotient.

As, the cost of golf ball box = $9.27.

So, 9.27 nearest to whole number is 9.

Thus to get the cost of each ball:

18 golf balls cost = $9.

So, 1 golf ball cost = $18\div 9=2.$

Therefore, each golf ball costs about $2.

3 0

3 years ago

I need some help with my math test today!

vladimir1956 [14]

Answer:

1.end point continues in one direction

2. point were two line segemnt, lines, rays meet

3. measures 90 degrees

4. greater than 90 degrees

5. measures 180 degrees

6. position in space

7. flat surface

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

PLS ANSWER ASAP 30 POINTS!!! CHECK PHOTO! WILL MARK BRAINLIEST TO WHO ANSWERS

Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

$a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\$

With respect to reference angle A, we have:

opposite side = a = 4
adjacent side = b = $\sqrt{33}$
hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

$\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\$

Then cosine which is adjacent over hypotenuse

$\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\$

Tangent is the ratio of opposite over adjacent

$\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\$

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

cosecant, abbreviated as csc, is the reciprocal of sine
secant, abbreviated as sec, is the reciprocal of cosine
cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

$\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{ ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

------------------------------------------------------

Summary:

The missing side is $b = \sqrt{33}$

The 6 trig functions have these results

$\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

7 0

1 year ago

I can't think of any right answer?​

I can't think of any right answer?