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3 years ago

Please answer the following questions !!!

Mathematics

2 answers:

dybincka [34]3 years ago

7 0

The first one is 5 the next one is 30.6 the third one is 6.4 the fourth one is 0.8 and the last one is 10.125

REY [17]3 years ago

5 0

1. 9 1/2
2. 30 3/5
3. 6 2/5
4. 1 1/5
5. 10 1/8

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Priya tried to solve an equation step by step. \qquad\begin{aligned} \dfrac f{0.25}&=16\\\\ \\ \dfrac{f}{0.25} \cdot0.25&amp

Zarrin [17]

Answer:

f=4

Step-by-step explanation:

Priya Tried to solve the equation below step-by-step:

$\qquad\begin{aligned} \dfrac f{0.25}&=16\\ \dfrac{f}{0.25} \cdot0.25&=16\cdot0.25&\green{\text{Step } 1}\\ f&=4&\blue{\text{Step } 2}\\\end{aligned}$

The steps are correct and indeed f=4.

7 0

3 years ago

Read 2 more answers

Priya filled 5 jars, using a total of 7 ½ cups of strawberry jam. How many cups of jam are in each jar?

Leya [2.2K]

Answer:

1 1/2 cups of jam per jar

Step-by-step explanation:

Divide 7 1/2 by 5 to get 1 1/2 cups of jam per jar.

6 0

3 years ago

Write the slope intercept form of the equation.

Elza [17]

Hello!

The slope intercept form equation is shown below.

y=mx+b

In this equation, m represents the slope and b is the y-intercept.

We can just plug in our numbers to find our answer.

y=3x-2

I hope this helps!

4 0

3 years ago

what about this one? is it correct? I am all done but would like to check if they are alright since I am so so rounding

Nutka1998 [239]

They're all correct! Nice job :)

7 0

3 years ago

The angle θ 1 is located in Quadrant IV, and cos ⁡ ( θ 1 ) = 9/ 19 , theta, start subscript, 1, end subscript, right parenthesis

Firdavs [7]

Answer:

$sin\theta_1 = - \frac{2\sqrt{70}}{19}$

Step-by-step explanation:

We are given that $\theta_1$ is in <em>fourth</em> quadrant.

$cos\theta_1$ is always positive in 4th quadrant and

$sin\theta_1$ is always negative in 4th quadrant.

Also, we know the following identity about $sin\theta$ and $cos\theta$ :

$sin^2\theta + cos^2\theta = 1$

Using \theta_1 in place of \theta:

$sin^2\theta_1 + cos^2\theta_1 = 1$

We are given that $cos\theta_1 = \frac{9}{19}$

$\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{280}{361}\\\Rightarrow sin\theta_1 = \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}$

$\theta_1$ is in <em>4th quadrant </em>so $sin\theta_1$ is negative.

So, value of $sin\theta_1 = - \frac{2\sqrt{70}}{19}$

6 0

3 years ago