Ask question

Ask question

All categories

3 years ago

15

6 - y = 12 is what lol

2 answers:

notsponge [240]3 years ago

7 0

Answer:

=−6

Step-by-step explanation:

zimovet [89]3 years ago

3 0

Answer:

y = -6

Step-by-step explanation:

You might be interested in

Which of the following is the number of sides for a regular polygon that will

marishachu [46]

Answer:

A

Step-by-step explanation:

A

4 0

3 years ago

A 10-foot ladder leans against a wall with its foot braced 3 feet from wall's base. How far up the wall does the ladder reach?

dlinn [17]

Pythagorean theorem
for a right triangle with legs legnth a and b and hytponuse c
a^2+b^2=c^2

the legnht of th eladder is the hypotnuse
the 3 feet is bottom leg
height is other leg

10=c
3=a
b=?
3^2+b^2=10^2
9+b^2=100
minus 9 both sides
b^2=91
sqrt both sides
b=√91
aprox
b=9.53939

answer is √91 feet or aprox 9.53939

8 0

3 years ago

242,628 written in word form

Svet_ta [14]

Two hundred forty two thousand, six hundred twenty eight.

3 0

3 years ago

Read 2 more answers

Enter an equation and an answer to the following question:

Naddik [55]

Answer:

$208.30

Step-by-step explanation:

<h2>Discount:</h2>

Let the original price be $ x

Discount % = 12%

Discount = $ 25

12% of x = 25

$\sf \dfrac{12}{100}*x = 25$

$\sf x = 25 * \dfrac{100}{12}=\dfrac{25*25}{3}\\$

= $ 208.30

6 0

2 years ago

Daniel [21]

Answer:

$x=\sqrt{\frac{7(4+\sqrt{15})}{2}}$

Step-by-step explanation:

From the way it is written, the $x$ is outside the square root. I will rewrite it as:

$x\sqrt{5} =x\sqrt{3} +\sqrt{7}$

$x\sqrt{5}-x\sqrt{3}=\sqrt{7}$

$x(\sqrt{5} - \sqrt{3} )=\sqrt{7}$

$x= \frac{\sqrt{7} }{\sqrt{5} - \sqrt{3}} \implies \frac{\sqrt{7}(\sqrt{5} + \sqrt{3}) }{2}$

$x=\frac{1}{2} \sqrt{7} (\sqrt{5} + \sqrt{3} )$

$x=\frac{\sqrt{35}}{2} +\frac{ \sqrt{21}}{2}$

$x=\frac{\sqrt{35}+\sqrt{21}}{2}$

Multiply denominator and numerator by 3

$x=\frac{3\sqrt{35}+3 \sqrt{21}}{6}$

Factor $\sqrt{3}$

$\sqrt{3} (\sqrt{105}+3 \sqrt{7})$

$x=\frac{\sqrt{3} (\sqrt{105}+3 \sqrt{7})}{6}$

Divide denominator and numerator by $\sqrt{3}$

$x=\frac{\sqrt{105}+3 \sqrt{7}}{2\sqrt{3} }$

Let's rewrite it again

$x=\frac{\sqrt{ (\sqrt{105}+3 \sqrt{7})^2}}{\sqrt{12} }$

$x=\sqrt{ \frac{1}{12} \cdot (\sqrt{105}+3 \sqrt{7})^2}$

It is already in the form $\sqrt{\frac{a}{b} }$

Expanding the perfect square, we have

$63+42\sqrt{15}+105$

$\frac{63}{12} +\frac{42\sqrt{15}}{12} +\frac{105}{12}$

$\frac{21}{4} +\frac{7\sqrt{15}}{2} +\frac{35}{4}$

Factor $\frac{7}{2}$

$\frac{7}{2} (4+\sqrt{15} )$

Therefore,

$x=\sqrt{\frac{7}{2} \left(4+\sqrt{15} \right)}$

$x=\sqrt{\frac{7(4+\sqrt{15})}{2}}$

7 0

3 years ago

Read 2 more answers