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3 years ago

7

I also bought $7$ different postcards. How many ways can I send the postcards to my $3$ friends, so that each friend gets at lea

st one postcard?
How do i count the restriction that each friend must get at least 1 or more postcards and the postcards are distinguishable?

1 answer:

harina [27]3 years ago

5 0

Bruh how’s it gonna tell me incorrect answer when it was right

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What is the fraction eight twentieths in its simplest form?

Nostrana [21]

Both the numerator and denominator are divisible by 4, meaning that when they are both divided by this number they wll be in their simplest form.

e.g. 8/4=2 and 20/4=5

Therefore the simplest form is 2/5

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3 years ago

Simplify 3x^2y^4 x 2x^6y

Feliz [49]

Answer:

6x^8y^5

Step-by-step explanation:

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add the exponents for the variables

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3 years ago

HELP QUICK!! solve by substitution and show work please

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Solve for the first variable in one of the equations, then substitute the result into the other equation.
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3 years ago

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

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

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

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3 years ago

marishachu [46]

4.85×88=$426.80.

1) $426.80 is the answer

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3 years ago