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max2010maxim [7]
3 years ago
13

When multiplying fractions, you multiply straight across? Yes or No ?

Mathematics
1 answer:
BaLLatris [955]3 years ago
8 0

Answer:

YES!

Step-by-step explanation:

You do multiply straight across for ex: 3/4 x 4/3= 12/12=1

For addition you don't but for multiplication you do!

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Imogene invested $8,000 in a bank account that pays 8 percent simple interest at the end of each year. Her friend invested the s
Natali5045456 [20]

The right system of equations to describe the situation would be on the form:

x1 = 8000 + y1*t

and

x2 = 8000 + y2*t

where x1 and x2 represents the total money of Imogene and her friend respectively at the end of t years.

Now for the value of amount earned, y1 and y2:

y1=8000*0.08

y2=2000*√(t-2)

<span> </span>

5 0
3 years ago
WHY DO I FEEL EXTRA NICE- FFTVJHFTYSDEV
enyata [817]

Answer:

cause u are nice

Step-by-step explanation:

5 0
3 years ago
In an atom, electrons have a negative charge and protons have a positive charge. One negative and
Ostrovityanka [42]

Answer:

The charge is negative and the amount is 215

Step-by-step explanation:

Electrons are negative

-354

Protons are positive

+139

-354+139

-215

6 0
3 years ago
If a current of 12 amps produces 480 volts across a resistor then how many volts does a 15 amp produce?​
LekaFEV [45]

Answer:

600 volts

Step-by-step explanation:

Given =

A = 12 ampere

V = 480 volts

A2 = 15 ampere

Solution =

R = V/I

= 480 / 12

= 40 ohms

R = V / I

V = R x I

= 40 x 15

= 600 volts

7 0
3 years ago
Can you help me with my work
andrezito [222]

Answer:

1. x = ±9

2. x=\pm \sqrt{13}

3. 12 and -12.

4. Antoine is incorrect. There exists two solutions x=5 and x= -5.

Step-by-step explanation:

According to the questions,

Problem 1. x^{2}-81=0 i.e. x^{2}=81 i.e. x = ±9.

Problem 2. 2x^{2}-26=0 i.e. x^{2}-13=0 i.e. x^{2}=13 i.e. x=\pm \sqrt{13}

Problem 3. [tex]f(x)=x^{2}-144[tex]

To find the roots, we take, [tex]x^{2}-144=0[tex] i.e. [tex]x^{2}=144[tex] i.e. x = ±12.

Thus, the options are 12 and -12.

Problem 4. We have [tex]f(x)=x^{2}+25[tex]

For the roots, we take, [tex]x^{2}+25=0[tex] i.e. [tex]x^{2}=25[tex] i.e. x = ±5.

Thus, Antoine is not correct and two solutions namely x=5 and x= -5 exists.

8 0
3 years ago
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