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creativ13 [48]
3 years ago
13

Is -1.5 greater than -3/2

Mathematics
2 answers:
Novosadov [1.4K]3 years ago
8 0
Yes -1.5 is greater than -3/2
Eduardwww [97]3 years ago
6 0
No, -1.5 is not greater than -3/2, they are equal.
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One half of the sum of 3 and q
ivann1987 [24]
Here you go let me know if you have questions
7 0
3 years ago
Can someone answer this? Thanks :)
Maslowich

Answer:

15ft

Step-by-step explanation:

Draw out the picture and use the Pythagorean theorem (equation is written in blue) to solve.

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4 0
3 years ago
I need help on number 2
Nina [5.8K]

9514 1404 393

Answer:

  a. 405

  b. -2

  c. 5

  d. 43

Step-by-step explanation:

I don't like doing arithmetic any more than you do, so I let a calculator do it for me. A spreadsheet works well for this, too. The function values are shown in the attachment.

Put the function argument where the variable is and do the arithmetic.

  a. h(4) = 5·3^4 = 405

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  c. h(0) = 5·3^0 = 5

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7 0
3 years ago
Find the approximate location of 2/3, 1/2, and 3/8 on a number line 0 to 1
schepotkina [342]

Answer:

0.6 (terminating), 0.5, 0.375

Step-by-step explanation:

I would assume you are wanting these in decimal form. In that case, you would have:

2/3 = 0.6 (terminating)

1/2 =  0.5

3/8 = 0.375

5 0
3 years ago
Read 2 more answers
Find the average value of the function on the given interval.<br> f(x)=2 ln x; [1,e]
JulijaS [17]

Answer:

f_{avg}=\frac{1}{e-1}

Step-by-step explanation:

We are given that a function

f(x)=2lnx

We have to find the average value of function on the given interval [1,e]

Average value of function on interval [a,b] is given by

\frac{1}{b-a}\int_{a}^{b}f(x)dx

Using the formula

f_{avg}=\frac{1}{e-1}\int_{1}^{e}lnx dx

By Parts integration formula

\int(uv)dx=u\int vdx-\int(\frac{du}{dx}\int vdx)dx

u=ln x and v=dx

Apply by parts integration

f_{avg}=\frac{1}{e-1}([xlnx]^{e}_{1}-\int_{1}^{e}(\frac{1}{x}\times xdx))

f_{avg}=\frac{1}{e-1}(elne-ln1-[x]^{e}_{1})

f_{avg}=\frac{1}{e-1}(e-0-e+1)=\frac{1}{e-1}

By using property lne=1,ln 1=0

f_{avg}=\frac{1}{e-1}

8 0
2 years ago
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