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3 years ago

7:5 is equivalent to...?

Mathematics

1 answer:

Varvara68 [4.7K]3 years ago

4 0

Answer:

6:0

Step-by-step explanation:

Thank me later :)))))))))

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Can someone help me with this

zmey [24]

Answer:

No, because they look like they are different sizes. Or you could say the first answer

3 0

3 years ago

Read 2 more answers

-4r-8(r-6)=7(5r-1)+8

KATRIN_1 [288]

Answer:

r = 1

Step-by-step explanation:

In this question, we have to solve for the vallue of "r". We will use the distributive property shown below to ease our process.

Distributive Property:

a(b+c) = ab + ac

Now, lets solve this:

$-4r-8(r-6)=7(5r-1)+8\\-4r-8r+48=35r-7+8\\-12r+48=35r+1\\48-1=35r+12r\\47=47r\\r=\frac{47}{47}\\r=1$

As we have seen the value of r is "1"

6 0

3 years ago

How many solutions can be found for the equation 4z 2(z − 4) = 3z 11? none one two infinitely many

Rus_ich [418]

There is only one solution for the equation 4z + 2(z -4) = 3z + 11 because the exponent for the power of z is 1.

<h3>What is an equation?</h3>

In mathematics, an equation is a formula that expresses the equality of two expressions, by connecting them with the equals sign =.

<h3>What is the Solution?</h3>

A solution is any value of a variable that makes the specified equation true.

According to the given information:

4z + 2(z-4)= 3z+11

Solve the equation,

4z+2z-8=3z+11

6z-3z=11+8

3z =19

Hence,

Number of solution that can be found for the equation 4z + 2(z-4)= 3z+11 is option(2) one

To know more about Equations and Solutions visit:

brainly.com/question/545403

#SPJ4

8 0

2 years ago

7 to the negative 3rd porwer times 7 to the negative 2nd porwer times 2 the 0 power divided by 7 to negative 1st power times 7 t

Jobisdone [24]

oof

remember some rules

(a/b)(c/d)=(ac)/(bd)

$\frac{x^a}{x^b}=x^{a-b}$

$x^0=1$

$(x^a)(x^b)=x^{a+b}$

$x^{-a}=\frac{1}{x^a}$

$\frac{(7^{-3})(7^{-2})(2^0)}{(7^{-1})(7^{-4})(2^3)}=$

$\frac{(7^{-3-2})(2^0)}{(7^{-1-4})(2^3)}=$

$\frac{(7^{-5})(2^0)}{(7^{-5})(2^3)}=$

$(\frac{7^{-5}}{7^{-5}})(\frac{2^0}{2^3})=$

$(7^{-5-(-5)})(2^{0-3})=$

$(7^0)(2^{-3})=$

$\frac{1}{2^3}=$

$\frac{1}{8}$

7 0

3 years ago

Find the area of the region that lies inside the first curve and outside the second curve. r = 6 − 6 sin θ, r = 6

yan [13]

Each curve completes one loop over the interval $0\le t\le2\pi$ . Find the intersections of the curves within this interval.

$6-6\sin\theta=6\implies 1-\sin\theta=1\implies \sin\theta=0\implies \theta=0,\theta=\pi$

The region of interest has an area given by the double integral

$\displaystyle\int_\pi^{2\pi}\int_6^{6-6\sin\theta}r\,\mathrm dr\,\mathrm d\theta$

equivalent to the single integral

$\displaystyle\frac12\int_\pi^{2\pi}\bigg((6-6\sin\theta)^2-6^2\bigg)\,\mathrm d\theta$

which evaluates to $9\pi+72$ .

8 0

3 years ago

7:5 is equivalent to...?​

7:5 is equivalent to...?