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3 years ago

PLS HELP, i am literally crying over this, Guaranteed brainliest if you get it right

Mathematics

2 answers:

katen-ka-za [31]3 years ago

5 0

Answer:

Step-by-step explanation:

14x+6/3=-8+4x

Using order of operations, this is how I would solve it

14x+6/3=-8+4x

14x+2=-8+4x

add 8 on both sides while subtracting 4x

10x+10=0

10x=-10

x=-1

Now, square -3 which is positive 9. 9-7 is 2. Hope I didn't make any mistakes because I'm a 6th grader btw. Hope it helps.

Agata [3.3K]3 years ago

4 0

I’m pretty sure the answer is -10

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Step-by-step explanation:

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250 ÷ 2 = 125

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Using intervals of width 10, how many intervals

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Step-by-step explanation:

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3 years ago

Solve the system substitution y= 4x + 1 and 3x+2y =13

strojnjashka [21]

Given :

${\qquad \sf \dashrightarrow y= 4x + 1 \: –––– \sf \: (i)}$

${\qquad \sf \dashrightarrow 3x + 2y = 13 \: –––– \sf \: (ii)}$

Now, Substituing the value of y in equation (ii) :

${\qquad \sf \dashrightarrow \: 3x + 2y = 13 }$

${\qquad \sf \dashrightarrow \: 3x + 2(4x + 1) = 13 }$

${\qquad \sf \dashrightarrow \: 3x + 8x + 2 = 13 }$

${\qquad \sf \dashrightarrow \: 11x + 2 = 13 }$

${\qquad \sf \dashrightarrow \: 11x = 13 - 2 }$

${\qquad \sf \dashrightarrow \: 11x = 11 }$

${\qquad \sf \dashrightarrow \: x = \dfrac{11}{11} }$

${\qquad \sf \dashrightarrow \bf \: x = 1 }$

Now, substituting the value of x in equation (i) :

${\qquad \sf \dashrightarrow y= 4x + 1}$

${\qquad \sf \dashrightarrow y= 4(1) + 1}$

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2 years ago

If 1/√a-√b=1/3 and 1/√a+√b=1/2, then find the difference of a and b.

kow [346]

ANSWER:

If $\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{1}{3}$ and $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{2}$ then the difference of a and b is 6

SOLUTION:

Given, $\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{1}{3}$ → $\sqrt{a}-\sqrt{b}=3$ ----- (1)

And $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{2}$ → $\sqrt{a}+\sqrt{b}=2$ --- (2)

We have to find difference of a and b.

Now, add (1) and (2)

$\sqrt{a}-\sqrt{b}=3$

$\sqrt{a}+\sqrt{b}=2$

Adding above two equations, we get,

$2 \sqrt{a}+0=2+3$

$\begin{array}{l}{2 \sqrt{a}=5} \\\\ {\sqrt{a}=\frac{5}{2}} \\\\ {a=\frac{25}{4}}\end{array}$

substitute $\sqrt{a}$ value in (2)

$\begin{array}{l}{\frac{5}{2}+\sqrt{b}=2} \\\\ {\sqrt{b}=\frac{2}{\sin \frac{5}{2}}} \\\\ {\sqrt{b}=\frac{4-5}{2}} \\\\ {\sqrt{b}=\frac{-1}{2}} \\\\ {b=\frac{1}{4}}\end{array}$

Now, difference of a and b is a – b = $\frac{25}{4}-\frac{1}{4}=\frac{24}{4}=6$

Hence, the difference of a and b is 6.

8 0

4 years ago