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3 years ago

U=a-k, solve for a. thanks for helping

Mathematics

1 answer:

Flauer [41]3 years ago

8 0

U=A-K
U+K=A
A=K+U THIS IS THE ANSWER

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Need help finding the value of x and y

kap26 [50]

Answer:

The values of x = 12 and y = 8.

Step-by-step explanation:

From the given figure ,

ΔMTW≅ΔBGK

That is, these two triangles are congruent.

If two triangles are congruent , all the corresponding angles and corresponding sides are equal.

Congruency is different from similarity . Similarity means two triangles which are the same with different dimensions.

Therefore , ∠MTW = ∠BGK

(4x - 3)° = 45°

4x = 48°

x = 12

Since ∠MTW = 45° ,

∠TMW = 180 - (41 +45)

= 180 - 86

=94°

From congruency ,

∠TMW = ∠GBK

94° = 11y + 6

11y = 88°

y = 8

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3 years ago

Five members of the soccer team and five members of the track team ran the 100-meter dash. Their times are listed in the table b

viva [34]

The last one....0.74

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3 years ago

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Someone please help me on this question ?:)

Whitepunk [10]

C , a relationship between two variables can be expressed by an equation in which one variable is equal to a constant.

8 0

3 years ago

If f(x)=3|x-2|, what is f(5.9)? a: 9 b: 10 c: 11 d: 12

Mice21 [21]

$|a|= \left\{\begin{array}{ccc}a&if&a\geq0\\-a&if&a \ \textless \ 0\end{array}\right\\\\f(x)=3|x-2|\\\\f(5.9)=3|5.9-2|=3|3.9|=3\cdot3.9=11.7$

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3 years ago

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Express the function as the sum of a power series by first using partial fractions. f(x) = 8 x2 − 4x − 12 f(x) = ∞ n = 0 find th

alexandr1967 [171]

I'm guessing the function is

$f(x)=\dfrac8{x^2-4x-12}=\dfrac8{(x-6)(x+2)}$

which, split into partial fractions, is equivalent to

$\dfrac1{x-6}-\dfrac1{x+2}$

Recall that for $|x|$ we have

$\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n$

With some rearranging, we find

$\dfrac1{x-6}=-\dfrac16\dfrac1{1-\frac x6}=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n$

valid for $\left|\dfrac x6\right|$ , or $|x|$ , and

$\dfrac1{x+2}=\dfrac12\dfrac1{1-\left(-\frac x2\right)}=\displaystyle\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n$

valid for $\left|-\dfrac x2\right|$ , or $|x|$ .

So we have

$f(x)=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n-\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n$

$f(x)=\displaystyle-\sum_{n=0}^\infty\left(\frac{x^n}{6^{n+1}}+\frac{(-x)^n}{2^{n+1}}\right)$

$f(x)=\displaystyle-\sum_{n=0}^\infty\frac{1+3(-3)^n}{6^{n+1}}x^n$

Taken together, the power series for $f(x)$ can only converge for $|x|$ , or $-2$ .

6 0

3 years ago