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2 years ago

9

Suppose both pairs of corresponding legs of two right triangles are congruent. Are the triangles congruent?

1 answer:

vredina [299]2 years ago

6 0

Answer:

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SOS if anyone could help that would be appreciated!!!❤️

Anastaziya [24]

Answer:

See below

Step-by-step explanation:

$2e)15-4x=10x+45$

Solve:

$15-4x=10x+45\\-4x-10x=45-15\\-14x=30\\ \boxed {x=-\dfrac{15}{7}}$ $3(x+1)=21$

Check:

$15-4 \left(-\dfrac{15}{7} \right)=10\left(-\dfrac{15}{7} \right)+45$

$15+\dfrac{60}{7} \right)=-\dfrac{150}{7} +45$

$\dfrac{210}{7} =30$

$30=30$

$3a) 3(x+1)=21$

Solve:

$3(x+1)=21\\3x+3=21\\3x=18\\x=6$

$3b) 2+3(x+1)=6x$

$2+3x+3=6x\\5=3x\\$

$x=\dfrac{5}{3}$

$3c) -2(2x-3)=-2$

Solve:

$-2(2x-3)=-2\\-4x+6=-2\\-4x=-8\\x=2$

3 0

3 years ago

What’s 2+8? A.10 B.6 C.387 D.11. Note: PLEASE HURRY IVE BEEN TRYING TO FIGURE THIS OUT ALL DAY!!

Leviafan [203]

Answer:

i think its 10 if not then 387

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it dive

Nostrana [21]

Answer:

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$ = 14.25

Step-by-step explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

$\sum_{k=0}^\infty a(r)^k$

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is $\sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}$

Given series,

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

$=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......$

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

Let

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$ and $t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

Now for $S_n$ ,

$S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......$

$=\sum_{n=0}^\infty9(\frac 17)^n$

It is a geometric series.

The common ratio of $S_n$ is $\frac17$

The sum of the series

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$

$=\frac{9}{1-\frac17}$

$=\frac{9}{\frac67}$

$=\frac{9\times 7}{6}$

=10.5

Now for $t_n$

$t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......$

$=\sum_{n=0}^\infty3(\frac 15)^n$

It is a geometric series.

The common ratio of $t_n$ is $\frac15$

The sum of the series

$t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

$=\frac{3}{1-\frac15}$

$=\frac{3}{\frac45}$

$=\frac{3\times 5}{4}$

=3.75

The sum of the series is $\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

= $S_n+t_n$

=10.5+3.75

=14.25

4 0

3 years ago

Matias preformed an operation with 7.6 and a number. He ended up moving the decimal point in 7.6 two places to the left. Matias

olga2289 [7]

Answer:

Matias divided 7.6 by 10 raised to the power of -2.

Step-by-step explanation:

Given : Matias preformed an operation with 7.6 and a number. He ended up moving the decimal point in 7.6 two places to the left.

To find : Matias divided 7.6 by 10 raised to the power of ?

Solution :

Matias preformed an operation with 7.6 and a number.

He ended up moving the decimal point in 7.6 two places to the left we have to multiply it with 0.01

i.e. $7.6\times 0.01=\frac{7.6}{100}=0.076$

So, Matias divided 7.6 by 10 raised to the power of -2.

5 0

3 years ago

What is .6 as a fraction

gladu [14]

0.6 as a fraction in simplest form is 3/5

3 0

3 years ago

Read 2 more answers