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3 years ago

Can someone one please give me the answer!!

Mathematics

1 answer:

melomori [17]3 years ago

5 0

So this is plotting and graphing basically. First you have to 1) draw the x and y axis table
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temp (°C)
2) Label the top of y axis (the line going up and down-vertical) dissolved oxygen content (ppm). The dependent variable- data collected will be graphed along this line

3) Label the side, x axis (the line that goes from left to right) temperature (°C). The independent variable- Is what I change, will be graphed along this line

4) I would recommend using two different colors to plot the difference in result on the graph....

5) On the graph lable x the temp 1-30 don't skip any #'s

6) On the y axis you're only aloud to plot whole numbers. I would do 1-14 and keep going
Hope this help :) I could draw it but I'm lazy.

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Phoenix [80]

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2 years ago

HELP How many pieces 3/7 of a foot each can I make out of 7/2 feet of rope?

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3 years ago

A rhombus and a square have one and the same side of 6 cm. The area of the rhombus is 4/5 of the area of the square. Find the he

Arada [10]

You will find the area the square and then find out what 4/5 of that area is.

A = bh
6 cm x 6 cm
A = 36 square cm

4/5 of 36 square cm
4/5 x 36
28 4/5 square cm

The area of the rhombus is 28 4/5 square cm. Use this to solve for the height of the rhombus.

A = bh
28 4/5 = 6 x h
6 6
h = 4 4/5 cm

The height of the rhombus is 4 4/5 cm.

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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 fourth 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 4th quadrant so $sin\theta_1$ is negative.

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

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2 years ago

John translated parallelogram ABCD using the rule (x,y)→(x+3, y-2). If angle A is 110° and angle B is 70°, what is the degree me

stiv31 [10]

This is kind of a trick question.
Translation is moving an object a certain distance. The original object and its translation have the same shape and size.
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3 years ago

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