Can someone help me with this one

adoni [48]

Answer:

$\huge\boxed{x=-4}$

Step-by-step explanation:

$3(x+9)-4x=11x-13x+23\\\\(3)(x)+(3)(9)-4x=(11x-13x)+23\\\\3x+27-4x=-2x+23\\\\(3x-4x)+27=-2x+23\\\\-x+27=-2x+23\qquad|\text{subtrsact 27 from both sides}\\\\-x=-2x-4\qquad|\text{add}\ 2x\ \text{to both sides}\\\\x=-4$

7 0

2 years ago

How many times would m^1/4 need to be expressed as a repeated factor in order to obtain m^2?

Marianna [84]

B.4
1/4 divided by 2
do the reciprocal
1/4 times 1/2=2/8
divide 8by 2

3 0

3 years ago

What is the common ratio for the geometric sequence? 32, 8, 2, 12, ... Enter your answer in the box.

mamaluj [8]

Answer:

Given sequence is not a geometric progression and there will be no common ration for this sequence.

Step-by-step explanation:

Need to determine common ration for the following geometric sequence

32, 8, 2, 12, ...

In given geometric sequence

a1 = 32, a2=8, a3=2, a4=12 ……….

Common ratio = $\frac{a_2}{a_1} =\frac{a_3}{a_2} =\frac{a_4}{a_3}$

$\frac{a_2}{a_1}=\frac{8}{32}=\frac{1}{4}$

$\frac{a_3}{a_2} =\frac{2}{8}=\frac{1}{4}$

$\frac{a_4}{a_3}=\frac{12}{2}=6$

Since $\frac{a_2}{a_1}=\frac{a_3}{a_2}\neq\frac{a_4}{a_3}$ so we can say that given sequence is not a geometric progression and there will no no common ration for this sequence.

3 0

3 years ago

What is the length of BC , rounded to the nearest tenth?

Arte-miy333 [17]

Step $1$

In the right triangle ADB

Find the length of the segment AB

Applying the Pythagorean Theorem

$AB^{2} =AD^{2}+BD^{2}$

we have

$AD=5\ units\\BD=12\ units$

substitute the values

$AB^{2}=5^{2}+12^{2}$

$AB^{2}=169$

$AB=13\ units$

Step $2$

In the right triangle ADB

Find the cosine of the angle BAD

we know that

$cos(BAD)=\frac{adjacent\ side }{hypotenuse}=\frac{AD}{AB}=\frac{5}{13}$

Step $3$

In the right triangle ABC

Find the length of the segment AC

we know that

$cos(BAC)=cos (BAD)=\frac{5}{13}$

$cos(BAC)=\frac{adjacent\ side }{hypotenuse}=\frac{AB}{AC}$

$\frac{5}{13}=\frac{AB}{AC}$

$\frac{5}{13}=\frac{13}{AC}$

solve for AC

$AC=(13*13)/5=33.8\ units$

Step $4$

Find the length of the segment DC

we know that

$DC=AC-AD$

we have

$AC=33.8\ units$

$AD=5\ units$

substitute the values

$DC=33.8\ units-5\ units$

$DC=28.8\ units$

Step $5$

Find the length of the segment BC

In the right triangle BDC

Applying the Pythagorean Theorem

$BC^{2} =BD^{2}+DC^{2}$

we have

$BD=12\ units\\DC=28.8\ units$

substitute the values

$BC^{2}=12^{2}+28.8^{2}$

$BC^{2}=973.44$

$BC=31.2\ units$

therefore

the answer is

$BC=31.2\ units$

8 0

3 years ago

Read 2 more answers

HELP HELP HELP PLS PLS PLS

Sergio [31]

Answer:

In order of the paragraph:

3, isosceles, greater, 3 in

Step-by-step explanation:

The first drop box has to be 3 because if you add up two sides of a triangle, the sum has to be greater than the third side. If two sides are equal, the triangle is isosceles. The third side cannot be 1 since 1+1 is not greater than 3. So the third side must measure 3in.

4 0

3 years ago