9. La mesure des côtés est représentée par une expression algébrique. Six

Help me again please due today

Mazyrski [523]

Answer:

y-int. is -4 (red)

Step-by-step explanation:

8 0

3 years ago

Neil wants to join three pieces of metal to form a right triangle. which of the following lengths could be used? please use step

vichka [17]

Answer: G. 12.5, 7.5, 10

Step-by-step explanation:

To determine the lengths of metal that could be used to form the tight angle triangle, we would apply Pythagoras theorem which is expressed as

Hypotenuse² = opposite side² + adjacent side²

The hypotenuse is the longest side while the opposite and adjacent sides are the shorter side. Therefore,

F.

8² = 4² + 4²

64 = 16 + 16 = 32

A right angle triangle cannot be formed because they are not equal.

G.

12.5² = 7.5² + 10²

156.25 = 56.25 + 100 = 156.25

A right angle triangle can be formed because they are equal.

H.

23² = 11² + 9²

529 = 121 + 81 = 202

A right angle triangle cannot be formed because they are not equal.

J.

96² = 12.5² + 6²

9216 = 156.25 + 36 = 192.25

A right angle triangle cannot be formed because they are not equal.

4 0

3 years ago

Jack recieved $600 for his birthday. If he saves it in a bank account with 8% interest compounded yearly, how much money will he

Gwar [14]

so 21 - 13, the money will be invested for 8 years.

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$600\\ r=rate\to 8\%\to \frac{8}{100}\to &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\to &1\\ t=years\to &8 \end{cases} \\\\\\ A=600\left(1+\frac{0.08}{1}\right)^{1\cdot 8}\implies A=600(1.08)^8\implies A\approx 1110.558126$

8 0

3 years ago

⦁ Write an equation of a line in slope intercept form (y = mx + b) with slope 1/3 going through the point (-6, 2). Show your wor

Zepler [3.9K]

$(\stackrel{x_1}{-6}~,~\stackrel{y_1}{2})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{2}=\stackrel{m}{\cfrac{1}{3}}(x-\stackrel{x_1}{(-6)}) \\\\\\ y-2=\cfrac{1}{3}(x+6)\implies y-2=\cfrac{1}{3}x+2\implies y=\cfrac{1}{3}x+4$

3 0

2 years ago

How many positive integers $n$ from 1 to 5000 satisfy the congruence $n \equiv 5 \pmod{12}$?

irga5000 [103]

The equivalence $n \equiv 5 \pmod{12}$

means that n-5 is a multiple of 12.

that is

n-5=12k, for some integer k

and so

n=12k+5

for k=-1, n=-12+5=-7

for k= 0, n=0+5=5 (the first positive integer n, is for k=0)

we solve 5000=12k+5 to find the last k

12k=5000-5=4995

k=4995/12=416.25

so check k = 415, 416, 417 to be sure we have the right k:

n=12k+5=12*415+5=4985

n=12k+5=12*416+5=4997

n=12k+5=12*417+5=5009

The last k which produces n<5000 is 416

For all k∈{0, 1, 2, 3, ....416}, n is a positive integer from 1 to 5000,

thus there are 417 integers n satisfying the congruence.

Answer: 417

6 0

3 years ago