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3 years ago

If 8 oranges cost $10.40, how many oranges can be bought for $33.80

Mathematics

2 answers:

love history [14]3 years ago

5 0

Answer:

24 if you can only buy it for every 8 bananas, but 26 if you can pay 1/8 of 10.40 for every banana

Step-by-step explanation:

MrMuchimi3 years ago

5 0

Oranges bought at ₹10.4 = 8
Oranges bought at ₹1 =8/₹10.4
Oranges bought at ₹33.8 = 8/₹10.4 × ₹33.8
= 26

Ans= 26 oranges

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hichkok12 [17]

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3 years ago

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Suppose you have a group of 10 children consisting of 4 girls and 6 boys. how many four -person teams can be chosen that consist

Ierofanga [76]

4 girls and 6 boys

2(2 girls) and 3(2 boys)

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4 years ago

Combine into a single logarithm.<br><br> 3log(x+y)+2log(x-y)-log(x^2 +y^2)

seropon [69]

Answer:

$3\log _{10}\left(x+y\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

Step-by-step explanation:

Given the expression

$3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)$

solving to write into a single logarithm

$3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$3\log _{10}\left(x+y\right)=\log _{10}\left(\left(x+y\right)^3\right)$

$=\log _{10}\left(\left(x+y\right)^3\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$2\log _{10}\left(x-y\right)=\log _{10}\left(\left(x-y\right)^2\right)$

$=\log _{10}\left(\left(x+y\right)^3\right)+\log _{10}\left(\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)$

$\log _{10}\left(\left(x+y\right)^3\right)+\log _{10}\left(\left(x-y\right)^2\right)=\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)$

$=\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)$

$\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

$=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

Thus,

$3\log _{10}\left(x+y\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

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3 years ago

There are 240 ways for cars to win first and second place in a race. How many cars are in the race?

melamori03 [73]

Answer:

120 cars

Step-by-step explanation:

Cause 240 and you have 1st and 2nd that's two places and 240/2 = 120.

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3 years ago

What are the solutions of the equation x squared equals 4? Use a graph of the related function.

Harman [31]

In the graph of quadratic equation, the parabola passes through the points left parenthesis (-2,0) right parenthesis, left parenthesis (0,-4) right parenthesis, and left parenthesis (2,0) right parenthesis.

<h3>What is a quadratic equation?</h3>

A quadratic equation is the equation in which the unknown variable is one and the highest power of the unknown variable is two.

The standard form of the quadratic equation is,

$ax^2+bx+c=0$

Here,(a,b, c) is the real numbers and (x) is the variable.

The given quadratic equation is,

$f(x)=x^2+4$

Equate the equation to zero to solve it further,

$x^2=4\\x^2-4=0\\x^2=4\\x=\sqrt{4}\\x=\pm2$

Thus, the solution of the equation is +2 and -2. The graph of the equation is attached below.

Hence, in the graph of quadratic equation, the parabola passes through the points left parenthesis (-2,0) right parenthesis, left parenthesis (0,-4) right parenthesis, and left parenthesis (2,0) right parenthesis.

Learn more about the quadratic equation here;

brainly.com/question/1214333

#SPJ1

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2 years ago

If 8 oranges cost $10.40, how many oranges can be bought for $33.80​

If 8 oranges cost $10.40, how many oranges can be bought for $33.80