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4 years ago

-6x-y=11 What’s the answer?

Mathematics

2 answers:

disa [49]4 years ago

4 0

0 is the correct answer!!! i hope this helps :))

Serhud [2]4 years ago

3 0

Answer:

_6x-y=11/ 6x+y+11=0 this is the answer

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Suppose you have abc with ab = 5, bc = 7, and ca = 10, and also efg with ef = 900, fg = 1260, and ge = 1800. are these triangles

madreJ [45]

SSS theorem: If two triangles have three pairs of sides in the same ratio, then the triangles are similar.

Write the ratio between the smallest sides, between the greatest sides and between the middle sides and compare the results:

$\dfrac{EF}{AB}=\dfrac{900}{5}=180;$
$\dfrac{EG}{AC}=\dfrac{1800}{10}=180;$
$\dfrac{GF}{CB}=\dfrac{1260}{7}=180.$

This gives you that

$\dfrac{EF}{AB}=\dfrac{EG}{AC}=\dfrac{GF}{CB}=180.$

By SSS theorem triangles ABC and EFG are similar with scale factor 180.

5 0

3 years ago

What is 42/134 as a percentage (with a explanation on how you caculated it)

BlackZzzverrR [31]

Answer:

31.34 (to 2d.p.)

Step-by-step explanation:

Just use a calculator and type 42/134

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3 years ago

Please help place the dots where the decimal numbers are on the number lines

Romashka [77]

Answer:

For the 0.6 put it on the 0.6

For the 0.43 put it closer to the 0.4

For the 5/100 put it on the 0.5

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3 years ago

sharla has 87 silver coins in her collection and 82 copper coins in her collection. She rounded each type of coin to the nearest

valina [46]

Answer:

no it is not reasonable

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4 years ago

An insurance company found that 9% of drivers were involved in a car accident last year. If seven drivers are randomly selected,

zzz [600]

Answer: 0.1061

Step-by-step explanation:

Given : An insurance company found that 9% of drivers were involved in a car accident last year.

Thus, the probability of drivers involved in car accident last year = 0.09

The formula of binomial distribution :-

$P(X=x)^nC_xp^x(1-p)^{n-x}$

If seven (n=7) drivers are randomly selected then , the probability that exactly two (x=2) of them were involved in a car accident last year is given by :-

$P(X=2)=^7C_2(0.09)^2(1-0.09)^{7-2}\\\\=\dfrac{7!}{2!5!}(0.09)^2(0.91)^{5}=0.106147867882\approx0.1061$

Hence, the required probability :-0.1061

3 0

3 years ago