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3 years ago

WILL MARK BRANLIEST AND THANK! 10 POINTS, EASY QUESTION!!

Mathematics

1 answer:

elena-14-01-66 [18.8K]3 years ago

4 0

Answer: 2±√3 and 1/(2±√3)

This might seem wrong, but you'll understand when you read the explanation.

Step-by-step explanation:

For this problem, you can write 2 equations to find the two numbers. Let's establish x for the first number and y for the second number.

Equation 1:

x+y=4

This equation comes from the problem that says the 2 numbers add up to 4.

Equation 2:

x*y=1

This equation comes from the part that says multiplies to 1.

Even if the problem states that the numbers can be negative, it is tricking us. We know in order for the product of 2 numbers to be positive, it has to be +*+ or -*-. Since the added numbers have to be +4, we know we cannot use -1 and 1 as our answer. What we can do is to use substitution to solve for x and y.

x+y=4 becomes y=4-x

We can substitute this into equation 2.

x(4-x)=1

4x-x²=1

4x-x²-1=0

-x²+4x-1=0

Since this equation is not factorable, we use the quadratic equation, $\frac{-b±\sqrt{b^2-4ac} }{2a}$ (<em>Please ignore the weird looking A that's in front of the ± sign. I don't know what's wrong with the system, but it's not supposed to be there)</em>

We plug our a, b, c into the equation and get x=2±√3.

Now that we know our x, we can plug it into any of the above equations.

(2±√3)(y)=1

y=1/(2±√3)

Our final 2 numbers are 2±√3 and 1/(2±√3)

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1. an alloy contains zinc and copper in the ratio of 7:9 find weight of copper of it had 31.5 kgs of zinc.

m_a_m_a [10]

Answer:

Step-by-step explanation:

Question (1). An alloy contains zinc and copper in the ratio of 7 : 9.

If the weight of an alloy = x kgs

Then weight of copper = $\frac{9}{7+9}\times (x)$

= $\frac{9}{16}\times (x)$

And the weight of zinc = $\frac{7}{7+9}\times (x)$

= $\frac{7}{16}\times (x)$

If the weight of zinc = 31.5 kg

31.5 = $\frac{7}{16}\times (x)$

x = $\frac{16\times 31.5}{7}$

x = 72 kgs

Therefore, weight of copper = $\frac{9}{16}\times (72)$

= 40.5 kgs

2). i). 2 : 3 = $\frac{2}{3}$

4 : 5 = $\frac{4}{5}$

Now we will equalize the denominators of each fraction to compare the ratios.

$\frac{2}{3}\times \frac{5}{5}$ = $\frac{10}{15}$

$\frac{4}{5}\times \frac{3}{3}=\frac{12}{15}$

Since, $\frac{12}{15}>\frac{10}{15}$

Therefore, 4 : 5 > 2 : 3

ii). 11 : 19 = $\frac{11}{19}$

19 : 21 = $\frac{19}{21}$

By equalizing denominators of the given fractions,

$\frac{11}{19}\times \frac{21}{21}=\frac{231}{399}$

And $\frac{19}{21}\times \frac{19}{19}=\frac{361}{399}$

Since, $\frac{361}{399}>\frac{231}{399}$

Therefore, 19 : 21 > 11 : 19

iii). $\frac{1}{2}:\frac{1}{3}=\frac{1}{2}\times \frac{3}{1}$

$=\frac{3}{2}$

$\frac{1}{3}:\frac{1}{4}=\frac{1}{3}\times \frac{4}{1}$

= $\frac{4}{3}$

Now we equalize the denominators of the fractions,

$\frac{3}{2}\times \frac{3}{3}=\frac{9}{6}$

And $\frac{4}{3}\times \frac{2}{2}=\frac{8}{6}$

Since $\frac{9}{6}>\frac{8}{6}$

Therefore, $\frac{1}{2}:\frac{1}{3}>\frac{1}{3}:\frac{1}{4}$ will be the answer.

IV). $1\frac{1}{5}:1\frac{1}{3}=\frac{6}{5}:\frac{4}{3}$

$=\frac{6}{5}\times \frac{3}{4}$

$=\frac{18}{20}$

$=\frac{9}{10}$

Similarly, $\frac{2}{5}:\frac{3}{2}=\frac{2}{5}\times \frac{2}{3}$

$=\frac{4}{15}$

By equalizing the denominators,

$\frac{9}{10}\times \frac{30}{30}=\frac{270}{300}$

Similarly, $\frac{4}{15}\times \frac{20}{20}=\frac{80}{300}$

Since $\frac{270}{300}>\frac{80}{300}$

Therefore, $1\frac{1}{5}:1\frac{1}{3}>\frac{2}{5}:\frac{3}{2}$

V). If a : b = 6 : 5

$\frac{a}{b}=\frac{6}{5}$

$=\frac{6}{5}\times \frac{2}{2}$

$=\frac{12}{10}$

And b : c = 10 : 9

$\frac{b}{c}=\frac{10}{9}$

Since a : b = 12 : 10

And b : c = 10 : 9

Since b = 10 is common in both the ratios,

Therefore, combined form of the ratios will be,

a : b : c = 12 : 10 : 9

7 0

3 years ago

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tatyana61 [14]

The equation of a cirecle with radius r and center at (h,k) is
(x-h)^2+(y-k)^2=r^2
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center=(-2,-4)
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(x+2)^2+(y+4)^2=100
(x,y)
sub given points (or use TI) and see if you get a true statmeent
A is on it

answer is A

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Answer:

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