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3 years ago

Identify the congruent criteria (mcqs) pls answer all of them

Mathematics

1 answer:

Contact [7]3 years ago

6 0

5. SSS
6. ASA
7. SAS
8. ASA
9. ASA

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How long will it take Greg to earn $500.00 round to the nearest hour

Mrrafil [7]

Answer:

56 hours

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $y/x=k$ or $y=kx$

Let

x -----> the number of hours

y ----> the amount earned in dollars

This problem represent a proportional relationship between the variable x and the variable y

Find the value of the constant of proportionality k

$k=y/x$

For (3,27) ----> $k=27/3=9\ \$/h$

For (9,81) ----> $k=81/9=9\ \$/h$

For (21,189) ----> $k=189/21=9\ \$/h$

The constant of proportionality k is 9

The linear equation is

$y=9x$

For y=$500

substitute in the equation and solve for x

$500=9x$

Divide by 9 both sides

$x=56\ hours$

5 0

3 years ago

Solving equations: (a) 5x - 9 = 3x + 3 (b) x + 6 = 34 - x

Jet001 [13]

Answer:

a) 5x - 9 = 3x + 3

5x - 3x = 3 + 9

2x = 12

x = 12 / 2

x = 6

b) x + 6 = 34 - x

x + x = 34 - 6

2x = 28

x = 28 / 2

x = 14

8 0

3 years ago

Read 2 more answers

⦁ Show (2.3)(5.06) as a fraction multiplication problem and explain why the answer is in thousandths (three decimal places).

astra-53 [7]

we have

$(2.3)(5.06)$

we know that

$2.3=\frac{23}{10}$

$5.06=\frac{506}{100}$

$(2.3)(5.06)=(\frac{23}{10})(\frac{506}{100})$

$=\frac{11,638}{1,000}$

$=11.638$

the answer is in thousandths, because the denominator of the multiplication of the two fractions is one thousand

8 0

3 years ago

Read 2 more answers

Madeline invests $25,000 in an account that offers a compound interest rate of 9.5% per year. Which of the following is the corr

LekaFEV [45]

Use this equation: Amount after years=Initial investment*(1+Interest rate/time compounded yearly)^number of years*times compounded yearly

So A=25,000(1+.095/1)^8*1
Simplify
A=25000(1.095)^8
Simplify
A=25000(2.07)
Solve
A=$51,671.73

This equation can be used for all problems of this type.

5 0

3 years ago

How do I find the integral ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

defon

$\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1$

$(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1$

$\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2$

$\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3$

$\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}$

4 0

3 years ago