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8090 [49]
3 years ago
7

Evaluate a+b for a =34 and b=-6

Mathematics
2 answers:
Digiron [165]3 years ago
5 0

Question

evaluate a+b for a =34 and b=-6

a + b =

34 + (-6) =

34 - 6 =

28

Artemon [7]3 years ago
5 0

You know that a and b should be plugged in the expression a+b

So: 34+(-6)

So: 34-6= 28

So your simplified answer will be 28.

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Svetllana [295]
\frac{16+x}{x^3}+\frac{7-4x}{x^3}=\frac{16+x+7-4x}{x^3}=\frac{23-3x}{x^3}\\\\======================================\\\\\frac{5}{t-1}+\frac{3}{t}=\frac{5t}{t(t-1)}+\frac{3(t-1)}{t(t-1)}=\frac{5t+3t-3}{t^2-t}=\frac{8t-3}{t^2-t}
7 0
3 years ago
Read 2 more answers
A triangle has coordinates A(6,1), B(6,5), and C(4,1). If the triangle is rotated 180 degrees, what are the new coordinates of t
4vir4ik [10]

Answer:

A(-6,-1) B(-6, -5) C(-4, -1)

Step-by-step explanation:

When a coordinate is rotated 180 degrees, they go from (x,y) to (-x,-y), so all you need to do is multiply every coordinate by -1

6 0
3 years ago
Suppose James randomly draws a card from a standard deck of 52 cards. He then places it back into the deck and draws a second ca
qwelly [4]

Answer:

1.92%

Step-by-step explanation:

The probability for first case, picking a queen out of deck, will be:

\frac{4}{52}

as there will be 4 queens in a deck, one of each suit.

For the second pick, the probability of picking a diamond card, will be:

\frac{13}{52}

here the total will remain 52 as he has replaced the first card and not kept it aside and there will be 13 cards in diamond suit (including the three face cards).

Thus the net probability for both cases will be:

P = \frac{4}{52}  * \frac{13}{52}\\ P = \frac{1}{52}\\ P = 0.01923\\P = 1.923\%\\\\P = 1.92\%

Thus total probability for the combined two cases will be 1.92%

7 0
3 years ago
Change the subject of the formula L = v 4kt - p to k.
Romashka [77]

Answer:

\boxed{k = \frac{L^2 + p}{4t}}

General Formulas and Concepts:

<u>Algebra I</u>

Basic Equality Properties

  1. Multiplication Property of Equality
  2. Division Property of Equality
  3. Addition Property of Equality
  4. Subtraction Property of Equality

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify given.</em>

<em />\displaystyle L = \sqrt{4kt - p}

<u>Step 2: Solve for </u><u><em>k</em></u>
We can use equality properties to help us rewrite the equation to get <em>k</em> as our subject:

Let's first <em>square both sides</em>:

\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow L^2 = \big( \sqrt{4kt - p} \big) ^2 \\& \rightarrow L^2 = 4kt - p \\\end{aligned}

Next, <em>add p to both sides</em>:

\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow L^2 = \big( \sqrt{4kt - p} \big) ^2 \\& \rightarrow L^2 = 4kt - p \\& \rightarrow L^2 + p = 4kt \\\end{aligned}

Next, <em>divide 4t by both sides</em>:

\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow L^2 = \big( \sqrt{4kt - p} \big) ^2 \\& \rightarrow L^2 = 4kt - p \\& \rightarrow L^2 + p = 4kt \\& \rightarrow \frac{L^2 + p}{4t} = k \\\end{aligned}

We can rewrite the new equation by swapping sides to obtain our final expression:

\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow \boxed{k = \frac{L^2 + p}{4t}}\end{aligned}

∴ we have <em>changed</em> the subject of the formula.

---

Learn more about Algebra I: brainly.com/question/27698547

---

Topic: Algebra I

6 0
2 years ago
Solve the following equations.<br> log2(x^2 − 16) − log^2(x − 4) = 1
Alenkasestr [34]

Answer:

x=\frac{4*(2+e)}{e-2}

Step-by-step explanation:

Let's rewrite the left side keeping in mind the next propierties:

log(\frac{1}{x} )=-log(x)

log(x*y)=log(x)+log(y)

Therefore:

log(2*(x^{2} -16))+log(\frac{1}{(x-4)^{2} })=1\\ log(\frac{2*(x^{2} -16)}{(x-4)^{2}})=1

Now, cancel logarithms by taking exp of both sides:

e^{log(\frac{2*(x^{2} -16)}{(x-4)^{2}})} =e^{1} \\\frac{2*(x^{2} -16)}{(x-4)^{2}}=e

Multiply both sides by (x-4)^{2} and using distributive propierty:

2x^{2} -32=16e-8ex+ex^{2}

Substract 16e-8ex+ex^{2} from both sides and factoring:

-(x-4)*(-8-4e-2x+ex)=0

Multiply both sides by -1:

(x-4)*(-8-4e-2x+ex)=0

Split into two equations:

x-4=0\hspace{3}or\hspace{3}-8-4e-2x+ex=0

Solving for x-4=0

Add 4 to both sides:

x=4

Solving for -8-4e-2x+ex=0

Collect in terms of x and add 4e+8 to both sides:

x(e-2)=4e+8

Divide both sides by e-2:

x=\frac{4*(2+e)}{e-2}

The solutions are:

x=4\hspace{3}or\hspace{3}x=\frac{4*(2+e)}{e-2}

If we evaluate x=4 in the original equation:

log(0)-log(0)=1

This is an absurd because log (x) is undefined for x\leq 0

If we evaluate x=\frac{4*(2+e)}{e-2} in the original equation:

log(2*((\frac{4e+8}{e-2})^2-16))-log((\frac{4e+8}{e-2}-4)^2)=1

Which is correct, therefore the solution is:

x=\frac{4*(2+e)}{e-2}

6 0
4 years ago
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