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2 years ago

Please help me, I'm stuck on this question.

Mathematics

2 answers:

ASHA 777 [7]2 years ago

8 0

Since the triangle is isosceles, then (x + 4) must be equal (3x - 8)

Solving the (x + 4) = (3x - 8), we'll get:

(x + 4) = (3x - 8)

3x - x = 8 + 4

2x = 12

x = 6

So, the length of AC is 6 units

never [62]2 years ago

6 0

Step-by-step explanation:

AC=AB+BC

AC=x+4+3x‐8

COMBINE LIKE TERMS

AC=x+3x

AC =4x+4‐8

=4x+‐4

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What is the 6th term of the geometric sequence where a1 = -4096 and a4 = 64?

Akimi4 [234]

$\bf \begin{array}{llccll}
term&value\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
a_1&-4096\\
a_2&-4096r\\
a_3&-4096rr\\
a_4&-4096rrr\\
&-4096r^3\\
&64
\end{array}\implies -4096r^3=64
\\\\\\
r^3=\cfrac{64}{-4096}\implies r^3=-\cfrac{1}{64}\implies r=\sqrt[3]{-\cfrac{1}{64}}
\\\\\\
r=\cfrac{\sqrt[3]{-1}}{\sqrt[3]{64}}\implies \boxed{r=\cfrac{-1}{4}}\\\\
-------------------------------$

$\bf n^{th}\textit{ term of a geometric sequence}\\\\
a_n=a_1\cdot r^{n-1}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
r=-\frac{1}{4}\\
a_1=-4096\\
n=6
\end{cases}
\\\\\\
a_6=-4096\left( -\frac{1}{4} \right)^{6-1}\implies a_6=-4^6\left( -\frac{1}{4} \right)^5$

4 0

2 years ago

Read 2 more answers

What's the value of x <br><br><br><br> (Middle School)

Elan Coil [88]

The answer is 360 hope it helps

6 0

2 years ago

Hi what is the answer for b please

Ray Of Light [21]

E = 352.50 euro × 1 pound / 1.41 euro = 250.00 pounds

J = 39856 yen × 1 pound / 188 yen = 212.00 pounds

a) Japan is cheaper

b) 250 - 212 = 38 pounds cheaper

7 0

3 years ago

How do I add and subtract radicals with no like terms

liq [111]

Most real number arithmetic is pretty vacuous

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

That's about as good a way as possible to write this particular real number. But it's far from the only way.

$(\sqrt 2 + \sqrt 3)^2 = \sqrt{2}^2 + 2\sqrt 2 \sqrt 3 + \sqrt{3}^2 = 5 + 2 \sqrt 6$

$\sqrt 2 + \sqrt 3 = \sqrt{5 + 2 \sqrt 6}$

Sometimes you can factor something out and you have a common radical:

$\sqrt{32} + \sqrt{128} = \sqrt{2^5} + \sqrt{2^7} = 4 \sqrt{2} + 8 \sqrt{2} = 12 \sqrt{2}$

But most of them are vacuous,

$3\sqrt{7} + 34 \sqrt{13} = 3\sqrt{7} + 34 \sqrt{13}$

8 0

3 years ago

The solution to an equation is

jok3333 [9.3K]

Answer:

solution is an assignment of values to the unknown variables that makes the equality in the equation true. In other words, a solution is a value or a collection of values (one for each unknown) such that, when substituted for the unknowns, the equation becomes an equality.

Step-by-step explanation:

I'm not really sure wym but this might help

5 0

2 years ago