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3 years ago

5

A recipe calls for 5 1/2 cups of flour. Lorenzo only has 3 1/4 cups of flour. Write and solve an equation to find the additional

amount of flour Lorenzo needs to make the recipe. (Solve Addition + Subtraction Equarions

1 answer:

anygoal [31]3 years ago

7 0

5 1/2 - 3 1/4 = f

That's a equation that you can solve.

Hope it helps! :D

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

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2 years ago

K + 9 = 17<br><br> What does k value?<br> k=8<br> k=9<br> k=17 <br> k=26

belka [17]

Answer:

8

Step-by-step explanation:

k + 9 = 17

subtract 9 from both sides:

k = 8

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3 years ago

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One week John earned three times as much as Alex earned mowing lawns. If they earned $48 altogether,how much did each one earn

Allisa [31]

Answer:Alex earned $12

John earned $36

Step-by-step explanation:

j=3a

a+3a=48

4a=48

a=48/4

a=12

j=36

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3 years ago

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denpristay [2]

Answer:

The first table

Step-by-step explanation: The x values don't repeat which makes it a function

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Diano4ka-milaya [45]

I’m Not good at This but I’ll try to help ;c

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