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2 years ago

If a area of a rectangle is 45... set up an equation that must be true: w = x L = x-4

Mathematics

1 answer:

Stella [2.4K]2 years ago

5 0

Answer: x times (x-4)=45

Step-by-step explanation:

First since the area is 45 that’s what the equation has to be equal to.
To find the area you always have to do length times width. So the width (x) times the length (x-4) equals 45.

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For this problem, involving a weighted die, assume that the outcomes 1 through 4 are equally likely, that the outcomes 5 through

storchak [24]

Answer:

a. $\frac{3}{14}$

b. $\frac{1}{14}$

Step-by-step explanation:

We are given that a die is rolled

Let x bet the cases in favor of 5.

Number of cases in favor of 6=x

Then according to question

Number of cases in favor of 1=3x

Number of cases in favor of 2=3x

Number of cases in favor of 3=3x

Number of cases in favor of 4=3x

Sum of total cases= $x+x+3x+3x+3x+3x=14x$

Probability of getting 5 = $\frac{x}{14x}=\frac{1}{14}$

Probability of getting 1= $\frac{3x}{14x}=\frac{3}{14}$

1.We have to find probability should be assigned to rolling a 4.

The probability of getting = $\frac{3}{14}$

2.We have to find the probability of rolling 6.

The probability of getting 6= $\frac{1}{14}$

3 0

3 years ago

-1/2 minus negative 5/9 equal what

Andrej [43]

I think it’s -19/18 hope it helped

8 0

3 years ago

Read 2 more answers

For the past three months, Grace used her cell phone for 43 minutes, 62 minutes, and 57 minutes how many minutes would she have

mina [271]

Answer:1

Step-by-step explanation:

Her average over those three months already is 54 so she has and average on 1 min left

8 0

3 years ago

Solve the following equations: (a) x^11=13 mod 35 (b) x^5=3 mod 64

tino4ka555 [31]

$x^{11}=13\pmod{35}\implies\begin{cases}x^{11}\equiv13\equiv3\pmod5\\x^{11}\equiv13\equiv6\pmod7\end{cases}$

By Fermat's little theorem, we have

$x^{11}\equiv (x^5)^2x\equiv x^3\equiv3\pmod5$

$x^{11}\equiv x^7x^4\equiv x^5\equiv6\pmod 7$

5 and 7 are both prime, so $\varphi(5)=4$ and $\varphi(7)=6$ . By Euler's theorem, we get

$x^4\equiv1\pmod5\implies x\equiv3^{-1}\equiv2\pmod5$

$x^6\equiv1\pmod7\impleis x\equiv6^{-1}\equiv6\pmod7$

Now we can use the Chinese remainder theorem to solve for $x$ . Start with

$x=2\cdot7+5\cdot6$

Taken mod 5, the second term vanishes and $14\equiv4\pmod5$ . Multiply by the inverse of 4 mod 5 (4), then by 2.

$x=2\cdot7\cdot4\cdot2+5\cdot6$

Taken mod 7, the first term vanishes and $30\equiv2\pmod7$ . Multiply by the inverse of 2 mod 7 (4), then by 6.

$x=2\cdot7\cdot4\cdot2+5\cdot6\cdot4\cdot6$

$\implies x\equiv832\pmod{5\cdot7}\implies\boxed{x\equiv27\pmod{35}}$

$x^5\equiv3\pmod{64}$

We have $\varphi(64)=32$ , so by Euler's theorem,

$x^{32}\equiv1\pmod{64}$

Now, raising both sides of the original congruence to the power of 6 gives

$x^{30}\equiv3^6\equiv729\equiv25\pmod{64}$

Then multiplying both sides by $x^2$ gives

$x^{32}\equiv25x^2\equiv1\pmod{64}$

so that $x^2$ is the inverse of 25 mod 64. To find this inverse, solve for $y$ in $25y\equiv1\pmod{64}$ . Using the Euclidean algorithm, we have

64 = 2*25 + 14

25 = 1*14 + 11

14 = 1*11 + 3

11 = 3*3 + 2

3 = 1*2 + 1

=> 1 = 9*64 - 23*25

so that $(-23)\cdot25\equiv1\pmod{64}\implies y=25^{-1}\equiv-23\equiv41\pmod{64}$ .

So we know

$25x^2\equiv1\pmod{64}\implies x^2\equiv41\pmod{64}$

Squaring both sides of this gives

$x^4\equiv1681\equiv17\pmod{64}$

and multiplying both sides by $x$ tells us

$x^5\equiv17x\equiv3\pmod{64}$

Use the Euclidean algorithm to solve for $x$ .

64 = 3*17 + 13

17 = 1*13 + 4

13 = 3*4 + 1

=> 1 = 4*64 - 15*17

so that $(-15)\cdot17\equiv1\pmod{64}\implies17^{-1}\equiv-15\equiv49\pmod{64}$ , and so $x\equiv147\pmod{64}\implies\boxed{x\equiv19\pmod{64}}$

5 0

3 years ago

SOMEONE PLEAAAAAAAAAAAASE HELP

rodikova [14]

4x+5 = 8x - 11
4x - 8x = -11 - 5
-4x = -16
x = 4

4x+5 = 4(4) +5 = 21
8x - 11 = 8(4) - 11 = 21

<RST = 21 + 21 = 42

answer
<RST = 42 degrees

6 0

3 years ago