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3 years ago

8

Solve the system of equations by substitution.

1 answer:

raketka [301]3 years ago

3 0

Answer:

in point form the answer is (32/3 , 6)

in equation form the answer is x=32/3 , y=6

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A square pyramid is sliced parallel to the base, as shown.

jeka94

There are 2 triangles once u slice in half so triangles is the right answer

8 0

3 years ago

Find the cost of leveling a badminton court 132m long and 58 m broad at the rate of Rs. 80 per sq. m.

serg [7]

Answer:

Rs. 612,480

Step-by-step explanation:

Hi there,

Here the question asks us to find the cost of leveling a badminton court at the rate of Rs.80 per sq..

So by the word leveling we know that we have to find the area of the court in order to proceed further.

<em>(I am assuming the court to be rectangular in shape).</em>

Area of a rectangle = Length * Breadth

*Length = 132 meters

*Breadth = 58 meters

==> 132*58= 7656 m^2

So now that we got the area of the badminton court lets find the cost of leveling it ==>

Cost of leveling per meter^2 = Rs. 80

Area = 7656 m^2

==> 7656 * 80 = 612,480

So the cost would be <u>Rs. 612,480</u>

<u></u>

<u></u>

<u><em>If you found this answer helpful please mark me as brainliest.</em></u>

8 0

2 years ago

Sort the ratios listed at the right into bins so that equivalent ratios are grouped together:

cestrela7 [59]

Step $1$

<u>Find the irreducible fraction in each ratio</u>

<u>case 1)</u> $\frac{40}{64}$

Divide by $8$ boths numerator and denominator

$\frac{40}{64}=\frac{5}{8}$

<u>case 2)</u> $\frac{5}{75}$

Divide by $5$ boths numerator and denominator

$\frac{5}{75}=\frac{1}{15}$

<u>case 3)</u> $\frac{14}{35}$

Divide by $7$ boths numerator and denominator

$\frac{14}{35}=\frac{2}{5}$

<u>case 4)</u> $\frac{24}{64}$

Divide by $8$ boths numerator and denominator

$\frac{24}{64}=\frac{3}{8}$

<u>case 5)</u> $\frac{6}{15}$

Divide by $3$ boths numerator and denominator

$\frac{6}{15}=\frac{2}{5}$

<u>case 6)</u> $\frac{65}{104}$

Divide by $13$ boths numerator and denominator

$\frac{65}{104}=\frac{5}{8}$

<u>case 7)</u> $\frac{66}{176}$

Divide by $22$ boths numerator and denominator

$\frac{66}{176}=\frac{3}{8}$

<u>case 8)</u> $\frac{12}{30}$

Divide by $6$ boths numerator and denominator

$\frac{12}{30}=\frac{2}{5}$

<u>case 9)</u> $\frac{15}{225}$

Divide by $15$ boths numerator and denominator

$\frac{15}{225}=\frac{1}{15}$

<u>case 10)</u> $\frac{6}{16}$

Divide by $2$ boths numerator and denominator

$\frac{6}{16}=\frac{3}{8}$

<u>case 11)</u> $\frac{15}{24}$

Divide by $3$ boths numerator and denominator

$\frac{15}{24}=\frac{5}{8}$

<u>case 12)</u> $\frac{48}{128}$

Divide by $16$ boths numerator and denominator

$\frac{48}{128}=\frac{3}{8}$

Step $2$

<u>Sort the ratios into bins</u>

1<u>) First Bin</u>

<u> $Ratio=\frac{5}{8}$ </u>

$\frac{40}{64}$

$\frac{65}{104}$

$\frac{15}{24}$

<u>2) Second Bin </u>

<u> $Ratio=\frac{1}{15}$ </u>

$\frac{5}{75}$

$\frac{15}{225}$

<u>3) Third Bin</u>

$Ratio=\frac{2}{5}$

$\frac{14}{35}$

$\frac{6}{15}$

$\frac{12}{30}$

4<u>) Fourth Bin</u>

<u> $Ratio=\frac{3}{8}$ </u>

$\frac{24}{64}$

$\frac{66}{176}$

$\frac{6}{16}$

$\frac{48}{128}$

3 0

3 years ago

Read 2 more answers

I need to know the improper fractions answers for:

zmey [24]

Answer:

Part 1) $x=\frac{15}{2}\ units$

Part 2) $z=\frac{15\sqrt{3}}{2}\ units$

Part 3) $y= \frac{15\sqrt{3}}{4}\ units$

Part 4) $b= \frac{45}{4}\ units$

Step-by-step explanation:

step 1

Find the value of x

In the large right triangle

$cos(60^o)=\frac{x}{15}$ ----> by CAH (adjacent side divided by the hypotenuse)

Remember that

$cos(60^o)=\frac{1}{2}$

substitute

$\frac{1}{2}=\frac{x}{15}$

solve for x

$x=\frac{15}{2}\ units$ ---> improper fraction

step 2

Find the value of z

In the large right triangle

Applying the Pythagorean Theorem

$15^2=x^2+z^2$

substitute the value of x

$15^2=(\frac{15}{2})^2+z^2$

solve for z

$z^2=15^2-(\frac{15}{2})^2$

$z^2=225-\frac{225}{4}$

$z^2=\frac{675}{4}$

$z=\frac{\sqrt{675}}{2}\ units$

simplify

$z=\frac{15\sqrt{3}}{2}\ units$

step 3

Find the value of y

In the right triangle of the right

$sin(30^o)=\frac{y}{z}$ ---> by SOH (opposite side divided by the hypotenuse)

substitute the given values of y and z

Remember that

$sin(30^o)=\frac{1}{2}$

so

$\frac{1}{2}=y:\frac{15\sqrt{3}}{2}$

solve for y

$\frac{1}{2}= \frac{2y}{15\sqrt{3}}$

$y= \frac{15\sqrt{3}}{4}\ units$

step 4

Find the value of b

In the right triangle of the right

$cos(30^o)=\frac{b}{z}$ ---> by CAH (adjacent side divided by the hypotenuse)

substitute the given values of y and z

Remember that

$cos(30^o)=\frac{\sqrt{3}}{2}$

so

$\frac{\sqrt{3}}{2}=b:\frac{15\sqrt{3}}{2}$

solve for y

$\frac{\sqrt{3}}{2}= \frac{2b}{15\sqrt{3}}$

$b= \frac{45}{4}\ units$

7 0

3 years ago

What’s is the mass of the triangle?

kakasveta [241]

Answer:

3 grams? i dunno

7 0

3 years ago

Read 2 more answers