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lara31 [8.8K]
2 years ago
9

You select a card from a standard deck of cards.

Mathematics
1 answer:
Shtirlitz [24]2 years ago
8 0

Answer:

0.3269

Step-by-step explanation:

Number of spades = 13

Number of 5's = 4

17/52 =

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How to change 7/10 to a percent
Ulleksa [173]

Answer:

70%

Step-by-step explanation:

since 7/10 is 0.7 as a decimal

you can then turn it into a percent by moving the decimal point two spots to the right

70 --> 70%

another way to do it is, to add zeros on both the numerator and denominator

then it is valued at 70/100

a percent is a number over 100

70 ---> 70%

your answer is 70%

hope this helps

7 0
2 years ago
Which ordered pair is a solution of the equation y equals x minus 3
Rom4ik [11]
Y = x-3

Basically you just need to plug in an x value and solve for y
y = x - 3
y = (4) - 3
y = 1

In this case, the ordered pair of this equation would be (4, 1)

You can do this given any x or y coordinate

Hope this helps 
6 0
3 years ago
How can you write the expression with rationalized denominator? 2+sqrt3(3)/sqrt3(6)
joja [24]
So we have a 6 at the bottom, and the root is 3, so hmm how to take it out, simple enough, just let's get something to make the 6 a 6³, so it comes out of the root

so 

\bf \cfrac{2+\sqrt[3]{3}}{\sqrt[3]{6}}\cdot \cfrac{\sqrt[3]{6^2}}{\sqrt[3]{6^2}}\implies \cfrac{(2+\sqrt[3]{3})(\sqrt[3]{6^2})}{(\sqrt[3]{6})(\sqrt[3]{6^2})}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{3}\cdot \sqrt[3]{36}}{\sqrt[3]{6^3}}
\\\\\\
\cfrac{2\sqrt[3]{36}+\sqrt[3]{3\cdot 36}}{6}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{108}}{6}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{3^3\cdot 4}}{6}
\\\\\\
\cfrac{2\sqrt[3]{36}+3\sqrt[3]{ 4}}{6}
8 0
3 years ago
Diana is planting a garden in the shape of the sector below. What is the length of fencing needed for the curred part of the gar
Lerok [7]

99/110

iruwjqkfsjakgjlkja;lkdfjskl ; jkgfj jh jj j hjh jh jh jh jhk kgoljkhk jhkl jh jkhjk k hjk hlkj h

3 0
3 years ago
Solving by elimination <br> 5x+2y= -3<br> 3x+3y=9<br> please help me solve x and y
-Dominant- [34]
1) -3(5x+2y=-3)⇒ -15x-6y=9
                                             ⇒ -9x=27
     2(3x+3y=9)⇒ 6x+6y=18

2) -9x/-9=27/-9 ⇒ x=-3 

3) 3(-3)+3y=9⇒ -9+9+3y=9+9⇒ 3y/3=18/3⇒ y=6

Answer: (-3,6)

Reasoning:
Step 1) In order to eliminate, first I had to multiple the first equation by -3 and the second by 2 so that when combining the equations y would cancel each other out so that I could solve for x. <em>Note: There are many combinations as to how you could multiple the equations so that either the x or y would cancel out.
</em>
Step 2) Once y is eliminated, solve for x.

Step 3) Now plug x back into one of the original equations and solve for y. <em>Note: Plug x back into one of the original equations, not the equations that were changed by multiplication,</em> 

6 0
3 years ago
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