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2 years ago

Insert one pair of bracket to make this calculation correct . 7 - 5 - 3 + 4 = 9

Mathematics

1 answer:

trasher [3.6K]2 years ago

3 0

7-(5-3)+4=9 is the answer

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Anthony jogs 11 1/2 miles in 23 mins. If Anthony continues to jog at a steady pace, how many miles will he jog each minute?

frozen [14]

For this case we can raise a rule of three based on Anthony trotting $11 \frac {1} {2}$ miles in 23 minutes.

$11 \frac {1} {2} = \frac {2 * 11 + 1} {2} = \frac {23} {2}$

$\frac {23} {2}$ miles -----------> 23 minutes

x ----------------------------------------> 1minute

$x = \frac {\frac {23} {2} * 1} {23}\\x = \frac {\frac {23} {2}} {23}\\x = \frac {23} {2 * 23}\\x = \frac {1} {2}$

ANswer:

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3 years ago

Find the value of x of the triangle

irina [24]

Answer:7

Step-by-step explanation:

It’s 7 their the same size as the other side of it

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2 years ago

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Help me plz The formula K=59(F−32)+273.15 converts temperatures from Fahrenheit F to Kelvin K.

quester [9]

Let's see what to do buddy...

Step-by-step explanation:

_________________________________

$k = 59(f - 32) + 273.15$

Subtract the sides of the equation minus -273.15

$k - 273.15 = 59(f - 32)$

Divided the sides of the equation by 59

$\frac{k - 273.15}{59} = \frac{59}{59} (f - 32) \\$

$\frac{k - 273.15}{59} = f - 32 \\$

Subtract the sides of the equation plus 32

$\frac{k - 273.15}{59} + 32 = f \\$

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

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3 years ago

1. What’s the exact value of cot[cot^-1 sqrt3/3] ?

vovikov84 [41]

So the problems ask to find and calculate the exact value of the trigonometric equation in the following equations and the best answers would be the following:
#1. sqrt(3)/3
#2. Arcsine of zero is 0
#3. x/sqrt(4-x^2)
I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications. Have a nice day

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Oliga [24]

On the bottom left, the answer is subtraction property of equality because you need to subtract 4x to the 7x to simplify the equation.

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2 years ago