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2 years ago

8-37 better than or equal to 29

Mathematics

1 answer:

Zina [86]2 years ago

5 0

The statement that 8-37 greater than or equal to 29 is false and the correct statement is 8 - 27 is less than 29

<h3>What is an inequality?</h3>

An inequality can be defined as a relation which makes an unequal comparison between two numbers or mathematical expressions.

It is used most often to compare two numbers on the number line by their size

The information given be represented as;

$8 - 37\geq 29$

Now, let's find the difference from the left side

We have;

$- 29\geq 29$

From this, we can see that the statement is wrong as - 29 is neither greater than or equal to 29

The correct statement should be; 8 - 27 is less than 29

Thus, the statement that 8-37 greater than or equal to 29 is false and the correct statement is 8 - 27 is less than 29

Learn more about inequalities here:

brainly.com/question/24372553

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Paula is located at a position of −12 feet relative to the ground. Which statement shows the distance she is from ground level?

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3 years ago

What is the length of if ?<br><br> 8 in.<br> 8.75 in.<br> 10.25 in.<br> 14 in.

Ket [755]

Answer:

A=8.75 in.

Step-by-step explanation:

Since ∆ HBA & ∆HKJ are similar,

HB/HA = HK/HJ

3/5.25 = 8/HJ

3HJ/5.25 = 8

3HJ = 42

HJ = 14 in.

AJ = 14 - 5.25

AJ = 8.75 in.

Therefore

A=8.75 in.

I hope this helped you.

5 0

4 years ago

Read 2 more answers

Find \tan\left(\frac{17\pi}{12}\right)tan( 12 17π )tangent, left parenthesis, start fraction, 17, pi, divided by, 12, end frac

uranmaximum [27]

One way to do this is to notice

$\dfrac{17\pi}{12}=\dfrac\pi6+\dfrac{5\pi}4$

Then

$\tan\dfrac{17\pi}{12}=\tan\left(\dfrac\pi6+\dfrac{5\pi}4\right)=\dfrac{\tan\frac\pi6+\tan\frac{5\pi}4}{1-\tan\frac\pi6\tan\frac{5\pi}4}$

We have

$\tan\dfrac\pi6=\dfrac{\sin\frac\pi6}{\cos\frac\pi6}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}$

and since $\tan x$ has a period of $\pi$ ,

$\tan\dfrac{5\pi}4=\tan\left(\pi+\dfrac\pi4\right)=\tan\dfrac\pi4=1$

and so

$\tan\dfrac{17\pi}{12}=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3$

7 0

3 years ago