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2 years ago

Question Which number has a 5 in the hundredths place?

Mathematics

2 answers:

Naily [24]2 years ago

7 0

Answer: This doesn't make sense is there a photo?

Step-by-step explanation:

AURORKA [14]2 years ago

6 0

Answer:

500

Step-by-step explanation:

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Simpson deposits $1,200 in a savings account that earns simple interest at the rate of 5% per year. What interest does he earn a

Verdich [7]

Answer:

$60

Step-by-step explanation:

1200*0.05 = $60

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2 years ago

Here is the histogram of a data distribution. All class widths are 1.

vaieri [72.5K]

The answer is 7 because the mean is an average. So if you add up all the numbers and then divide that number by the amount of numbers listed then that will get you your answer which is 7

4+5+6+7+8+9+10=49
49/7=7

4 0

2 years ago

Analyzing a Graphical Relationship

Step2247 [10]

Answer: A : decreasing B : constant C: increasing D: Constant

Step-by-step explanation:

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3 years ago

Jamal states that ax + b = a(x + c), given a, b, and c are not equal to 0. What must be the value of c for Jamal's statement to

Anna35 [415]

ANSWER

For Jamal statement to be true,

$c = \frac{b}{a}$

EXPLANATION

The given statement is:

ax + b = a(x + c)

We expand the left hand side to obtain,

ax + b = ax + ac

Subtract ax from from both sides of the equation.

b=ac

We know that, a≠0 , so we can divide through by 'a'.

$\frac{b}{a} = c$

6 0

3 years ago

Find the exact value of each trigonometric function for the given angle θ.

Kay [80]

Answer:

$\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=\dfrac{2}{\sqrt{3}}.$

Step-by-step explanation:

The given angle is 240 degrees.

We need to find the exact value of each trigonometric function for the given angle θ.

Since $\theta=240$ , it means θ lies in 3rd quadrant. In 3d quadrant only tan and cot are positive.

$\sin (240^\circ)=\sin (180^\circ+60^\circ)=-\sin (60^\circ)=-\dfrac{\sqrt{3}}{2}$

$\cos (240^\circ)=\cos (180^\circ+60^\circ)=-\cos (60^\circ)=-\dfrac{1}{2}$

$\tan (240^\circ)=\tan (180^\circ+60^\circ)=\tan (60^\circ)=\sqrt{3}$

$\cot (240^\circ)=\cot (180^\circ+60^\circ)=\cot (60^\circ)=\dfrac{1}{\sqrt{3}}$

$\sec (240^\circ)=\sec (180^\circ+60^\circ)=-\sec (60^\circ)=-2$

$\csc (240^\circ)=\csc (180^\circ+60^\circ)=-\csc (60^\circ)=-\dfrac{2}{\sqrt{3}}$

Therefore, $\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=-\dfrac{2}{\sqrt{3}}.$

8 0

2 years ago