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3 years ago

Find two consecutive whole numbers that square root 97 lies between

Mathematics

2 answers:

lora16 [44]3 years ago

8 0

Answer: 9 and 10

Step-by-step explanation:

To understand this question we need to first understand how to solve square root equations! The easiest way to learn how to solve square root problems is by finding the reverse of the problem.

For example, if you tried to find the square root of 25, you would reverse the problem to do 5^2 = 5 x 5 = 25.

So if we were to implement this problem, we would look at these two whole numbers.

9^2 = 9 x 9 = 81

10^2 = 10 x 10 = 100

Thus meaning that the only two numbers that the square root of 97 lies between can be 9 and 10!

It's important to rely on your knowledge of multiplication to develop a strong base for square roots.

antoniya [11.8K]3 years ago

7 0

Answer:

9 & 10

Step-by-step explanation:

97 lies between two perfect squares, 81 and 100

Hence sqrt(97) would lie between sqrt(81) and sqrt(100), which are 9 and 10

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Which expression uses the associative property to make it easier to evalute?

mixas84 [53]

Answer:

The correct option is (c).

Step-by-step explanation:

The given expression is :

$14(\dfrac{3}{2}\times \dfrac{1}{4})$

We need to use the associative property to make it easier.

The associative property for multiplication is as follows :

$A\times (B\times C)=(A\times B)\times C$

We have,

$A=14, B=\dfrac{3}{2}\ and\ C=\dfrac{1}{4}$

$14(\dfrac{3}{2}\times \dfrac{1}{4})=(14\times \dfrac{3}{2})\times \dfrac{1}{4}$

Hence, the correct option is (c).

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2 years ago

James spent $(2y2 + 6). He bought notebooks that each cost $(y2 − 1).

CaHeK987 [17]

The expression to find the number of notebooks James bought would be written as:

Number of notebooks = total spent / cost per item

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If y = 3, then the number of notebooks bought would be:

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3 years ago

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$tan(\alpha)=cot(\beta)$

Find the value of the length side adjacent to the angle alpha

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

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