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3 years ago

What is 1 4/5 + 1 1/5??

Mathematics

1 answer:

Rudik [331]3 years ago

8 0

$1\dfrac{4}{5}+1\dfrac{1}{5}=(1+1)+\left(\dfrac{4}{5}+\dfrac{1}{5}\right)=2+\dfrac{1+4}{5}=2+\dfrac{5}{5}=2+1=3$

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Why is standard form best for factoring an equation? I need a reason why please!

BabaBlast [244]

Step-by-step explanation:

Be sure that your equation is in standard form (ax2+bx+c=0) before you start your factoring attempt

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3 years ago

Solve the inequality X+ 7 < 4x+ 25 The solution is x

Mazyrski [523]

Answer:

$x > -6$

Step-by-step explanation:

Solve the following inequality:

$x + 7 < 4x + 25$

-Take $4x$ and subtract it from $x$ :

$x - 4x + 7 < 4x - 4x + 25$

$-3x + 7 < 25$

-Subtract $7$ on both sides:

$-3x + 7 - 7< 25 - 7$

$-3x < 18$

-When you are dividing an integer by a negative integer, then the inequality sign changes. So, Take $-3x$ and divide it on both sides:

$\frac{-3x}{-3} < \frac{18}{-3}$

$x > -6$

Therefore, the answer is $x > -6$ .

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3 years ago

Work out the problem for 17/20 +7/20

notka56 [123]

Answer:

6/5

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

a baker has 6 small bags of flour.Eaxh bag weighs 1 pound . she divides each bag into thirds . How many 1/3 pound bags of flour

Naddika [18.5K]

If she had 6 bags before and each bag became three, the total number is 6*3=18 bags.

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4 years ago

Read 2 more answers

Margee thinks she can use logs to solve 56 = x^8, since logs seem to make exponents disappear. Unfortunately, Margee is wrong. E

aniked [119]

Answer:

We use log to solve only those equations in which we have our variables in power form.

Step-by-step explanation:

Given : Margee thinks she can use logs to solve $56=x^8$ , since logs seem to make exponents disappear but Margee is wrong

We have to explain the difference between equations like $2=(1.04)^x$ and $56=x^8$

We use log to solve only those equations in which we have our variables in power form.

Out of given equation only $2=(1.04)^x$ has x in power form so we can apply log for solving the equation as,

$2=(1.04)^x\\\\\ \text{Taking ln both sides},\\\ln (2)=\ln (1.04)^x\\\\\\\\text{Using idenity} \ln a^b=b\ln a\\\\\\ln (2)=\ln (1.04)^x\\\\\text{on solving, we get}\\\\x=17.673$

While solving other equation ,

$56=x^8$ , we can directly take 8 root both side, $56=x^8\\\\\\56^{\frac{1}{8}}=x^{(8\times{\frac{1}{8}})$

Thus, We use log to solve only those equations in which we have our variables in power form.

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3 years ago