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wel
3 years ago
9

The first and last terms in a proportion are called the amar

Mathematics
2 answers:
Ilya [14]3 years ago
5 0

Answer:

What is a Proportion? The four numbers a, b, c and d are known as the terms of a proportion. The first a and the last term d are referred to as extreme terms while the second and third terms in a proportional are called mean terms.

Step-by-step explanation:

Dafna1 [17]3 years ago
4 0

Answer:

False

Step-by-step explanation:

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Y > x + 1
Romashka-Z-Leto [24]

Answer:A D F

Step-by-step explanation:

6 0
2 years ago
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How can proportional reasoning help solve a problem?
Anestetic [448]
Answer: Proportional reasoning compares ratios to answer questions.
6 0
3 years ago
There are 3 3/4 pounds of bricks in a bag. Each brick weighs 5/8 of a pound. How many bricks are in the bag.
Bess [88]

Answer: There are 6 bricks in the bag.

Step-by-step explanation:

Convert from mixed number to decimal number. To do it, divide the numerator of the fraction by the denominator and add the result to the whole number part. Then:

3\frac{3}{4}\ lb=(3+0.75)\ lb=3.75\ lb

Convert from fraction to decimal number. To do it you need to divide the numerator by the denominator. Then, you get:

\frac{5}{8}lb=0.625\ lb

Let be "x" the number of bricks in the bag<em>.</em>

<em> </em>Based on the information given in the exercise, you can set up the following proportion:

\frac{1}{0.625}=\frac{x}{3.75}

Finally, you must solve for "x" in order to find its value. This is:

(3.75)(\frac{1}{0.625})=x\\\\x=6

8 0
3 years ago
1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a
katovenus [111]

1. Let a and b be coefficients such that

\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}

Combining the fractions on the right gives

\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}

\implies 1 = (2a+b)x + 3a

\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23

so that

\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}

2. a. The given ODE is separable as

x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}

Using the result of part (1), integrating both sides gives

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C

Given that y = 1 when x = 1, we find

\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)

so the particular solution to the ODE is

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)

We can solve this explicitly for y :

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)

\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|

\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|

\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}

2. b. When x = 9, we get

y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}

8 0
2 years ago
Find the absolute value of lk-8l=8<br><br> PLS ANSWER BEFORE 6:00 TODAY TY
Bingel [31]

It's 7:58 here.

|k-8|=8

k-8 = 8 or -(k-8)=8

k=16 or -k+8=8

k=16 or k=0

check:

|0-8|=|-8|=8 good

|16-8|=|8|=8 good

Answer: k=16 or k=0

5 0
3 years ago
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