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3 years ago

True or false: the square root of a number is always less than that number. Select one: A. true B. false

Mathematics

2 answers:

aleksandrvk [35]3 years ago

8 0

Answer:

True

Step-by-step explanation:

allochka39001 [22]3 years ago

4 0

Answer:

True

Step-by-step explanation:

Say your trying to find the square root of 400 hundred it would be 20 because 20 time 20 is 400..it will always be smaller

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Write the value of 83,479

Ann [662]

Eighty three thousand four hundred and seventy nine

6 0

3 years ago

Does the graph represent a function? y 1 6 5+ 4+ 3+ 2 1+ +++ -7 -6 -5 -4 -3 -2 tot 2 1 2 3 4 5 6 7 -2+ -3 -4+ -5- -67 -7 Choose

Deffense [45]

Yes the graph does represent a function

8 0

2 years ago

Read 2 more answers

Please help me thanks :)

Lina20 [59]

Answer:

Step-by-step explanation:

To sovle this we must combine the numbers togther and the variables togther.

We have:

8c+6-3c-2

Lets take out the c's and combine them:

8c -3c

Now lets take out the numbers:

6-2

Now lets put these back together and we get the expression:

5c+4

This looks like A.

Hope this helps! :D

8 0

3 years ago

Could anyone help me with this problem i'm stuck <br><br> h(17)= ?

nikdorinn [45]

Answer:

$\sqrt{17t}$

Step-by-step explanation:

Consider the right side of the table

It informs us of the values for h(t) for different values of t

The top one informs us what h(t) is when t = 17, that is

h(t) = $\sqrt{17t}$ , when t = 17, thus

h(17) = $\sqrt{17t}$

4 0

3 years ago

Help I am very bad at math I need answers by tomorrow if you can help please do

d1i1m1o1n [39]

Answer:

Step-by-step explanation:

(a - b)(a +b) = a² - b²

1 - Sin² A = Cos² A

$LHS = \frac{1}{1- Sin A} + \frac{1}{1 + Sin A}\\\\= \frac{1*(1 + Sin A)}{(1- Sin A)(1 + Sin A)} + \frac{1*(1- Sin A)}{(1 + Sin A)(1- Sin A)}\\\\= \frac{1 + Sin A+ 1 - Sin A}{1^{2}- Sin^{2} A}\\\\= \frac{2}{1 - Sin^{2} A}\\\\= \frac{2}{Cos^{2} A}\\\\= 2 Sec^{2} A$

2) Sec² A - Tan² A = 1

$LHS = \frac{1}{Sec A - Tan A}\\\\=\frac{1*(Sec A + Tan A)}{(Sec A - Tan A)(Sec A + Tan A)}\\\\=\frac{Sec A + Tan A}{Sec^{2} A - Tan^{2} A}\\\\=\frac{Sec A + Tan A }{1}\\\\= Sec A + Tan A = RHS\\\\\\$

3) LHS = Cosec² A + Cot² A

= Cosec² A + Cosec² A - 1

= 2Cosec² A - 1 = RHS

$4) LHS = \frac{Sec A}{Cos A}- \frac{Tan A}{Cot A}\\\\ = Sec A*\frac{1}{Cos A}-Tan A*\frac{1}{Cot A}\\\\ = Sec A * Sec A - Tan A * Tan A\\\\= Sec^{2} A - Tan^{2} A \\\\= 1$

3 0

3 years ago