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3 years ago

0.5y+4=7 can some one please help me with this problem it will be great.

Mathematics

2 answers:

Brut [27]3 years ago

7 0

Answer:

If u have to find the value of y it is y=6

Step-by-step explanation:

Rainbow [258]3 years ago

5 0

Answer:

Step-by-step explanation:

0.5y+4=7

0.5y=7-4

0.5÷0.5y=3÷0.5

y=6

( if they want tha value of y)

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A wedding planner uses 72 ivy stems for 18 centerpieces when she arrives at the venue she realizes she will only need 16 centerp

beks73 [17]

Answer:

64 ivy stems

Step-by-step explanation:

To solve this problem we can use rule of three:

Firstly we had 72 ivy stems to use in 18 centerpieces, and know we have 16 centerpieces and want to know how many ivy stems we need to use, so:

18 centerpieces -> 72 ivy stems

16 centerpieces -> X ivy stems

18/16 = 72/X

X = 16*72/18 = 64

He needs to use now 64 ivy stems to maintain the ratio.

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3 years ago

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Determine and solve the system of equations. A parking meter takes nickels and dimes only. Ninety-five coins are removed from th

kvv77 [185]

Answer:

$4x=3+2ysquare over 3$

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Step-by-step explanation:

7 0

3 years ago

21-7x=7(3-x)<br> Solve this please!!!!!!!!!!

Hunter-Best [27]

Answer:

All real numbers

Step-by-step explanation:

21 - 7x = 7(3 - x)

21 - 7x = 21 - 7x

Since both sides of the equation are equal, the answer is all real numbers.

3 0

3 years ago

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What is the difference between the two graphs at X = -3

schepotkina [342]

Answer:

Step-by-step explanation:

Blue: when x = - 3, y = 5

Green: when x = -3, y = 0

The difference between the two graphs at X = -3 : 5 - 0 = 5

Answer

6 0

2 years ago

Given that −4i is a zero, factor the following polynomial function completely. Use the Conjugate Roots Theorem, if applicable. f

GalinKa [24]

Answer:

$\large \boxed{\sf \bf \ \ f(x)=(x-4i)(x+4i)(x+3)(x-5) \ \ }$

Step-by-step explanation:

Hello, the Conjugate Roots Theorem states that if a complex number is a zero of real polynomial its conjugate is a zero too. It means that (x-4i)(x+4i) are factors of f(x).

$\text{Meaning that } (x-4i)(x+4i) =x^2-(4i)^2=x^2+16 \text{ is a factor of f(x).}$

The coefficient of the leading term is 1 and the constant term is -240 = 16 * (-15), so we a re looking for a real number such that.

$f(x)=x^4-2x^3+x^2-32x-240\\\\ =(x^2+16)(x^2+ax-15)\\\\ =x^4+ax^3-15x^2+16x^2+16ax-240$

We identify the coefficients for the like terms, it comes

a = -2 and 16a = -32 (which is equivalent). So, we can write in $\mathbb{R}$ .

$\\f(x)=(x^2+16)(x^2-2x-15)$

The sum of the zeroes is 2=5-3 and their product is -15=-3*5, so we can factorise by (x-5)(x+3), which gives.

$f(x)=(x^2+16)(x^2-2x-15)\\\\=(x^2+16)(x^2+3x-5x-15)\\\\=(x^2+16)(x(x+3)-5(x+3))\\\\=\boxed{(x^2+16)(x+3)(x-5)}$

And we can write in $\mathbb{C}$

$f(x)=\boxed{(x-4i)(x+4i)(x+3)(x-5)}$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

7 0

2 years ago