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3 years ago

8

James Works in a flower shop. He will put 36 tulips in vases for a wedding. He must use the same number of tulips in each vase.H

ow many tulips could be in each vase ?

1 answer:

soldi70 [24.7K]3 years ago

5 0

Answer:

1,2,3,4,6,9,12,18,36

Step-by-step explanation:

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Alex

233.3333333. and you cant sell .333 of a book so 234 books

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3 years ago

A right triangle as shown below

anastassius [24]

Answer:

sin a = 7/25

cos a = 24/25

tan a = 7/24

Step-by-step explanation:

Trig. How wonderful. I get tripped up on these types of problems some times, so I decided to try to help! To start, write out the three ratios.

SOH (sine=opposite/hypotenuse) CAH (cosine=adjacent/hypotenuse) TOA (tangent=opposite/adjacent)

Then, label the triangle with “hypotenuse” “adjacent” and “opposite.” This helps us correctly use and find the raitos. Then, use these ratios to find out the ratios of A!

sin a = 7/25

cos a = 24/25

tan a = 7/24

If needed, just divide the ratios to get their decimal form!

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3 years ago

What SPC tool best indicates the shape of the data population and indicates if there are any gaps in the data

romanna [79]

The gaps in the area needs to be closed so the data you use is lenient

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2 years ago

Answer? with solution please. Will automatically report kung wala:)

serious [3.7K]

Answer:

B

Step-by-step explanation:

Communitive only works with multiplication and addition. It states that if you have something adding or subtracting something switched around, they are the same. B has that so the answer is B

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3 years ago

Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)

DENIUS [597]

Compute the necessary values/derivatives of $f(x)$ at $x=1$ :

$f(1)=3$

$f'(1)=2$

$f''(1)=-12$

$f'''(1)=-12$

$f^{(4)}(1)=48$

$f^{(5)}(1)=120$

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) $f(x)$ at $x=1$ by

$T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5$

$T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5$

###

Another way of doing this would be to solve for the coefficients $a,b,c,d,e,g$ in

$f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5$

by expanding the right hand side and matching up terms with the same power of $x$ .

5 0

3 years ago