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3 years ago

-2.3x + 4.2x = 15.01 Solve for x

Mathematics

2 answers:

Neko [114]3 years ago

6 0

-2.3x + 4.2x = 15.01 original equation

1.9x = 15.01 combine like terms

1.9x/1.9 = 15.01/1.9 divide both sides by 1.9 to get x alone

x = 7.9 simplify

CHECK YOUR WORK:

plug in the solution to x into the original equation like so:

-2.3(7.9) + 4.2(7.9) = 15.01 original equation w/ substitute for x

-18.17 + 33.18 = 15.01 simplify multiplication

15.01 = 15.01 simplify addition

When you get an answer like this, you know that the solution to x that you found is correct.

VashaNatasha [74]3 years ago

4 0

Answer:

x = 7.9

Step-by-step explanation:

-2.3x + 4.2x = 15.01

combine like terms

1.9x = 15.01

divide both sides by 1.9

x = 7.9

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The temperature at sunrise was -3*F. By noon, the temperature climbed 18 degrees. By sundown the temperature was -14*F. How much

Natalija [7]

Answer:

32*F

Step-by-step explanation:

* = degrees

Started at -3*F

By noon, the temperature climbed 18* = +18*

By sundown the temperature was -14*F

What we want to know: How much did the temperature drop between noon and sundown?

This is actually really easy!

What was the temperature at noon? --> 18*F

What was the temperature at sundown? --> -14*F

So what is the difference between 18* and -14*? We can't use an equation, so we will solve manually.

The distance from -14 to 0 is 14 degrees. Keep that in mind.

The distance from 18 to 0 is 18 degrees. Keep that in mind.

So what is 14 + 18? 14 + 18 = 32*.

So, the temperature dropped 32*F between noon and sundown.

I hope this helps! I know this was a little messy, but that is because we couldn't use an equation/expression to solve this.

<h2>PLEASE MARK BRAINLIEST!</h2>

6 0

3 years ago

Round to the nearest tenth.

sammy [17]

Answer:

$c=9.3\ units$

Step-by-step explanation:

step 1

Find the measure of angle A

we know that

The sum of the interior angles in any triangle must be equal to 180 degrees

$A+B+C=180^o$

substitute the given values

$A+110^o+30^o=180^o\\A=40^o$

step 2

Find the length side c

Applying the law of sines

$\frac{a}{sin(A)}=\frac{c}{sin(C)}$

substitute the given values

$\frac{12}{sin(40^o)}=\frac{c}{sin(30^o)}$

solve for c

$c=\frac{12}{sin(40^o)}sin(30^o)$

$c=9.3\ units$

4 0

4 years ago

You roll two number cubes.

Brums [2.3K]

Answer:

Its independent, because the outcome of the first roll doesnt affect the outcome of the second roll.

Step-by-step explanation:

6 0

4 years ago

Read 2 more answers

An element with a mass 310 grams decays by 5.7% per minute. How much of the element is remaining after 9 minutes, to the nearest

balu736 [363]

Mass of an element = 310 grams

Percentage of the element that will decay in a minute = 5.7%

Quantity of the element that will decay in one minute :

$= \frac{5.7 \times 310}{100}$

$= \frac{1767}{100}$

$= \tt17.67 \: grams$

Thus, 17.67 grams of the element will decay every minute.

Quantity of element that will decay after nine minutes :

$= 17.67 \times 9$

$=\tt 159.03 \: grams$

Thus, 159.03 grams of the element will decay after 9 minutes.

Quantity of element that will remain after 9 minutes :

$= 310 - 159.03$

$= \tt150.97 \: grams$

Thus, 150.97 grams of the element will remain after nine minutes.

We can round off 150.97 to 151.00.

Therefore, 151.00 grams of the element will remain after 9 minutes.

6 0

3 years ago

Please explain your answer :D don’t answer if you will not provide an explanation please :”)

aleksley [76]

Answer:

Step-by-step explanation:

QUESTION 7:

We are trying to find angles 1 and 2.

Known angle: 107˚.

In this question, we have 2 parallel lines with a transversal. A transversal is a line that passes through 2 parallel lines making it so that the angles on both sides are exactly the same.

Step 1: Find the measures of all the angles using the one we already know.

I will start with the angle adjacent (right next to) the angle that measures 107˚. As you can see, it is a supplementary angle because both of them form a straight line, or 180˚.

Therefore,

107+x= 180

x= 180-107

x= 73˚

Next, I will find the measure of the angle that is vertical to the one that measures 107˚. Note that angles that are vertical to each other are always identical, or in other words they have the same measure. Hence, the angle "vertical", or opposite to the one that measures 107˚ has to measure 107˚ as well. You can also tell this is true because they look the same.

Now I'll find the measure of the angle vertical to the one that measures 73˚. As we know, vertical angles are identical, so the one vertical to it is 73˚.

Here, I'll solve for angles 1 and 2. Like I said earlier, when a transversal crosses through 2 parallel lines, there will be 2 sets of angles that all have the same measurements.

To solve for angle 1, simply know what angle is right below it, and that is the one that measures 107˚. (If you look at the 2 lines, you will see that both angle sets are essentially duplicates of each other.) Likewise, angle 2 has to be 73˚ because the one under it is also 73˚.

Angle 1: 107˚

Angle 2: 73˚

QUESTION 8:

Again, we have a transversal that passes through 2 parallel lines. One angle measures 95˚. When there are 3 seperate angles along a straight line, the one next to the one with the given measurement will be the same as that one.

So angle 3 is 95˚. To find angle 4, find the angle measurement that's vertical to angle 3, which is 95˚. Now, since that angle and angle 4 are supplementary (both form a straight line which is always 180˚), 95+x= 180.

(x stands for angle 4)

x+95= 180

x= 180-95

x= 85

Angle 4= 85˚.

QUESTION 9:

ANGLE 5: 49˚

When there is a transversal, there are 2 identical pairs of angles. Therefore, angle 5 must be 49˚.

ANGLE 6: 131˚

angle 5 and angle 6 form a straight line which means they are supplementary angles. Therefore, both angles must add up to 180˚. So 49+x= 180

x= 180-49= 131

(x stands for angle 6)

Hope this helps!! Also I worked pretty hard on this so brainliest would be appreciated :D

Practice problems if it helps :) : https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-geometry/cc-8th-angles-between-lines/e/parallel_lines_1

4 0

2 years ago