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Harlamova29_29 [7]
3 years ago
5

1/a+2/b+3/c; Express as a single fraction.

Mathematics
2 answers:
Anika [276]3 years ago
8 0
Answer: (bc+2ac+3ab)/(abc)
UNO [17]3 years ago
8 0

For this case we must express as a single fraction the following expression:\frac {1} {a} + \frac {2} {b} + \frac {3} {c}

We have that the common denominator of the three fractions is given by: abc

So, to express as a fraction we divide abc by each denominator and multiply by each numerator, obtaining:

\frac {bc + 2ac + 3ab} {abc}

Answer:

\frac {bc + 2ac + 3ab} {abc}

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What would be the answer please explain
sp2606 [1]

Using the line of the best fit, the predicted student's score in English test is 48

<h3>How to determine the student's score in English?</h3>

From the question, we have:

Mathematics score = 60

The scores in English tests are plotted on the y-axis.

On the given graph, we have:

(x,y) = (60,48)

This means that when x = 60, the value of y is 48

This in other words means that the student's score in English test is 48

Read more about line of best fit at:

brainly.com/question/17261411

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6 0
2 years ago
Which of the following is a point where the graph
ser-zykov [4K]

Answer:

-When (2×-5) intersects at ×-axis the value of × will be zero.

=>F(×)=× 0=2×-5

hence,

0,=2×-5

5=2×

=>×=5/2

Step-by-step explanation:

hpe it hlps

7 0
3 years ago
In the diagram, the measure of 1 is 55°. If line y is rotated 5° clockwise about the point of
REY [17]
A. 60 degrees is the answer
5 0
2 years ago
Read 2 more answers
PLEASE HELP ME WITH THIS ONE ASAP THANKS (will mark brainlest)
jeka57 [31]
The answer is 8000 meters I think
4 0
3 years ago
Find a polynomial P(x) with real coefficients having a degree 4, leading coefficient 4, and zeros 3-i and 4i.
Alisiya [41]

now, this polynomial has roots of 3-i and 4i, namely 3 - i and 0 + 4i.

let's bear in mind that a complex root never comes all by her lonesome, her sibling is always with her, the conjugate, so if 3 - i is there, 3 + i is also coming along, likewise if 0 + 4i is there, her sibling 0 - 4i is also there.

\bf \begin{cases} x=3-i\implies &x-3+i=0\\ x=3+i\implies &x-3-i=0\\ x=4i\implies &x-4i=0\\ x=-4i\implies &x+4i=0 \end{cases}\\\\[-0.35em] ~\dotfill\\\\ (x-3+i)(x-3-i)(x-4i)(x+4i)=\stackrel{y}{0} \\[2em] \underset{\textit{difference of squares}}{[(x-3)+i][(x-3)-i]}\underset{\textit{difference of squares}}{[x-4i][x+4i]}=0

\bf [(x-3)^2-i^2][x^2-(4i)^2]=y\implies [(x-3)^2-(-1)][x^2-(4^2i^2)]=0 \\[2em] [(x-3)^2-(-1)][x^2-(16(-1))]=0\implies [(x-3)^2+1][x^2+16]=0 \\[2em] [(x^2-6x+9)+1][x^2+16]=y\implies (x^2-6x+10)(x^2+16)=0 \\\\\\ x^4-6x^3+10x^2+16x^2-96x+160=0 \\\\\\ x^4-6x^3+26x^2-96x+160=0 \\\\\\ \stackrel{\textit{multiplying both sides by 4}}{4(x^4-6x^3+26x^2-96x+160)=4(0)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 4x^4-24x^3+104x^2-384x+640=y~\hfill

3 0
3 years ago
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