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3 years ago

WHOEVER GETS IT RIGHT GETS BRAINLIEST SHOW YOUR WORK OR ELSE YOU WILL BE BLOCKED AND REPORTED

Mathematics

1 answer:

Anna71 [15]3 years ago

6 0

Hi,
The answer is 4 sq. . cm,
If you see closely, it is nine times smaller, so we divide 36/9=4
So the area of the smaller figure is 4 sq. cm.
Hope this helps you.

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Hi. Can someone pls help

Marina CMI [18]

(2,10) this is because 7 x 2 is 14, minus 4 that’s 10

8 0

3 years ago

Read 2 more answers

A new cellular phone tower services all phones within a 20 mile radius. Doreen lives 16 miles east and 12 miles south of the tow

Lunna [17]

Answer: Yes.

Step-by-step explanation:

Hi, since the situation forms a right triangle (see attachment) we have to apply the Pythagorean Theorem:

c^2 = a^2 + b^2

Where c is the hypotenuse of the triangle (in this case the distance between Doreen’s house and the tower) and a and b are the other sides.

Replacing with the values given:

c^2 = 16^2 + 12^2

c^2 = 256+144

c^2 = 400

c = √400

c = 20 miles

So, she is within the area serviced by the tower (20 miles)

Feel free to ask for more if needed or if you did not understand something.

3 0

4 years ago

Which set of numbers shows all factors of 18

Alexxx [7]

Answer:

Step-by-step explanation: Factors of 18: The square root of 18 is 4.2426, rounded down to the closest whole number is 4. Testing the integer values 1 through 4 for division into 18 with a 0 remainder we get these factor pairs: (1 and 18), (2 and 9), (3 and 6). The factors of 18 are 1, 2, 3, 6, 9, 18.

hope this helps ^-^

5 0

3 years ago

Help me to answer now ineed this <br> Please...

Vera_Pavlovna [14]

ANSWER TO QUESTION 1

$\frac{\frac{y^2-4}{x^2-9}} {\frac{y-2}{x+3}}$

Let us change middle bar to division sign.

$\frac{y^2-4}{x^2-9}\div \frac{y-2}{x+3}$

We now multiply with the reciprocal of the second fraction

$\frac{y^2-4}{x^2-9}\times \frac{x+3}{y-2}$

We factor the first fraction using difference of two squares.

$\frac{(y-2)(y+2)}{(x-3)(x+3)}\times \frac{x+3}{y-2}$

We cancel common factors.

$\frac{(y+2)}{(x-3)}\times \frac{1}{1}$

This simplifies to

$\frac{(y+2)}{(x-3)}$

ANSWER TO QUESTION 2

$\frac{1+\frac{1}{x}} {\frac{2}{x+3}-\frac{1}{x+2}}$

We change the middle bar to the division sign

$(1+\frac{1}{x}) \div (\frac{2}{x+3}-\frac{1}{x+2})$

We collect LCM to obtain

$(\frac{x+1}{x})\div \frac{2(x+2)-1(x+3)}{(x+3)(x+2)}$

We expand and simplify to obtain,

$(\frac{x+1}{x})\div \frac{2x+4-x-3}{(x+3)(x+2)}$

$(\frac{x+1}{x})\div \frac{x+1}{(x+3)(x+2)}$

We now multiply with the reciprocal,

$(\frac{(x+1)}{x})\times \frac{(x+2)(x+3)}{(x+1)}$

We cancel out common factors to obtain;

$(\frac{1}{x})\times \frac{(x+2)(x+3)}{1}$

This simplifies to;

$\frac{(x+2)(x+3)}{x}$

ANSWER TO QUESTION 3

$\frac{\frac{a-b}{a+b}} {\frac{a+b}{a-b}}$

We rewrite the above expression to obtain;

$\frac{a-b}{a+b}\div {\frac{a+b}{a-b}}$

We now multiply by the reciprocal,

$\frac{a-b}{a+b}\times {\frac{a-b}{a+b}}$

We multiply out to get,

$\frac{(a-b)^2}{(a+b)^2}$

ANSWER T0 QUESTION 4

To solve the equation,

$\frac{m}{m+1} +\frac{5}{m-1} =1$

We multiply through by the LCM of $(m+1)(m-1)$

$(m+1)(m-1) \times \frac{m}{m+1} + (m+1)(m-1) \times \frac{5}{m-1} =(m+1)(m-1) \times 1$

This gives us,

$(m-1) \times m + (m+1) \times 5}=(m+1)(m-1)$

$m^2-m+ 5m+5=m^2-1$

This simplifies to;

$4m-5=-1$

$4m=-1-5$

$4m=-6$

$\Rightarrow m=-\frac{6}{4}$

$\Rightarrow m=-\frac{3}{2}$

ANSWER TO QUESTION 5

$\frac{3}{5x}+ \frac{7}{2x}=1$

We multiply through with the LCM of $10x$

$10x \times \frac{3}{5x}+10x \times \frac{7}{2x}=10x \times1$

We simplify to get,

$2 \times 3+5 \times 7=10x$

$6+35=10x$

$41=10x$

$x=\frac{41}{10}$

$x=4\frac{1}{10}$

Method 1: Simplifying the expression as it is.

$\frac{\frac{3}{4}+\frac{1}{5}}{\frac{5}{8}+\frac{3}{10}}$

We find the LCM of the fractions in the numerator and those in the denominator separately.

$\frac{\frac{5\times 3+ 4\times 1}{20}}{\frac{(5\times 5+3\times 4)}{40}}$

We simplify further to get,

$\frac{\frac{15+ 4}{20}}{\frac{25+12}{40}}$

$\frac{\frac{19}{20}}{\frac{37}{40}}$

With this method numerator divides(cancels) numerator and denominator divides (cancels) denominator

$\frac{\frac{19}{1}}{\frac{37}{2}}$

Also, a denominator in the denominator multiplies a numerator in the numerator of the original fraction while a numerator in the denominator multiplies a denominator in the numerator of the original fraction.

That is;

$\frac{19\times 2}{1\times 37}$

This simplifies to

$\frac{38}{37}$

Method 2: Changing the middle bar to a normal division sign.

$(\frac{3}{4}+\frac{1}{5})\div (\frac{5}{8}+\frac{3}{10})$

We find the LCM of the fractions in the numerator and those in the denominator separately.

$(\frac{5\times 3+ 4\times 1}{20})\div (\frac{(5\times 5+3\times 4)}{40})$

We simplify further to get,

$(\frac{15+ 4}{20})\div (\frac{(25+12)}{40})$

$\frac{19}{20}\div \frac{(37)}{40}$

We now multiply by the reciprocal,

$\frac{19}{20}\times \frac{40}{37}$

$\frac{19}{1}\times \frac{2}{37}$

$\frac{38}{37}$

5 0

3 years ago

A publisher reports that 45% of their readers own a laptop. A marketing executive wants to test the claim that the percentage is

nexus9112 [7]

Answer:

$z=\frac{0.40 -0.45}{\sqrt{\frac{0.45(1-0.45)}{370}}}=-1.933$

$p_v =2*P(z$

Step-by-step explanation:

Information given

n=370 represent the sample selected

$\hat p=0.4$ estimated proportion of readers owned a laptop

$p_o=0.45$ is the value that we want to test

z would represent the statistic

$p_v$ represent the p value

Creating the hypothesis

We need to conduct a hypothesis in order to test if the true proportion of readers owned a laptop is different from 0.45, the system of hypothesis are:

Null hypothesis: $p=0.45$

Alternative hypothesis: $p \neq 0.45$

The statistic is:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing we got:

$z=\frac{0.40 -0.45}{\sqrt{\frac{0.45(1-0.45)}{370}}}=-1.933$

Calculating the p value

We have a bilateral test so then the p value would be:

$p_v =2*P(z$

8 0

3 years ago