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jasenka [17]
3 years ago
12

What is 3/8 + 1/2 + 1/4 ​

Mathematics
1 answer:
Anni [7]3 years ago
6 0

Answer:

9/8

Step-by-step explanation:

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A classmate simplifies the expression in exercise 1. is the expression he wrote equivalent to the original expression? explain.
AlladinOne [14]
Excercise 1:

No, this is not simplified fully. The full answer would be <span>−7dk+14d−21k</span>−7, not <span>14d - 9 - 21k - 7dk + 2. So, it is not equivalent.
</span><span>
Excercise 4:
</span><span>7dk (2 - 3 - 1) - 7
</span>
Just multiply 7dk into (2) (-3) and (-1)
you'd get:
<span><span>−<span>14dk</span></span>+</span>−<span>7 after simplifying it fully.
</span>
p = parameter
w = width

w+2.75

p = 2w+2(w+2.75) = <span>4w</span>+<span>5.5 = 30

Answer to the last one:

w = 6.125</span>
4 0
3 years ago
If possible, find AB. &amp; State the dimension of the result.
Nikitich [7]

Answer:

Step-by-step explanation:

A=\begin{bmatrix}0 &0  &5 \\ 0 &0 &-3 \\ 0 &0  &3 \end{bmatrix}

B=\begin{bmatrix}8 &-12  &5 \\ 7 &19 &5 \\ 0 &0  &0 \end{bmatrix}

A.B = A × B

A.B=\begin{bmatrix}0 &0  &0 \\ 0 &0  &0 \\ 0 &0  &0 \end{bmatrix}

Dimension of the resultant matrix is (3 × 3)

3 0
3 years ago
select the values that make the inequality h/4 greater than or equal to -1 true then write an equivalent inequality in terms of
jok3333 [9.3K]

Answer:

Values that make it equal:

-4, -3, -1, 0, 4

Written equivalent inequality:

h≥−4

Step by Step for the written inequality:  

h/4 ≥−1

Multiply both sides by 4. Since 4 is positive, the inequality direction remains the same.

h≥−4

I hope this helps you!! :)

3 0
3 years ago
Find the ratio if it is given that ...
igomit [66]
Here is the answer to your question as This is the way I think it is solved

4 0
3 years ago
Read 2 more answers
Find the area of the shaded region.f(x)=4x+3x2−x3,g(x)=0<br> The area is _____.
Tasya [4]

Answer:

The answer is " \bold{\frac{7}{2}}"

Step-by-step explanation:

Given value:

\to f(x)=4x+3x^2-x^3\\\\\to g(x)=0

Find:

area=?

calculation:

\ Area = \int_{-1}^{0} -(4x+3x^2-x^3) dx  + \int_{0}^{1} (4x+3x^2-x^3) dx  \\\\

        =  -(2x^2+x^3- \frac{x^4}{4})_{-1}^{0} + (2x^2+x^3- \frac{x^4}{4})_{0}^{1}\\\\\\=  -( 2x^2+x^3- \frac{x^4}{4})_{-1}^{0} + (2x^2+x^3- \frac{x^4}{4})_{0}^{1}\\\\\\= ( 2- 1 -\frac{1}{4}) + (2+1- \frac{1}{4})\\\\\\=  ( \frac{8-4-1}{4}) + (\frac{8+4-1}{4})\\\\=  ( \frac{3}{4}) + (\frac{11}{4})\\\\=   \frac{3}{4} + \frac{11}{4}\\\\= \frac{11+3}{4}\\\\= \frac{14}{4}\\\\= \frac{7}{2}\\\\

7 0
3 years ago
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