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3 years ago

Which of the following is an equation of the line through (-4, 12) parallel to the line y= - 4x - 4?

Mathematics

1 answer:

Virty [35]3 years ago

8 0

Step-by-step explanation:

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Determine the domain of the following graph:

Rama09 [41]

Answer:

Step-by-step explanation:

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3 years ago

Mr. Toshiro manages a company that supplies a variety of domestic and imported nuts to supermarkets. He received an order for 12

alexandr1967 [171]

Answer:

n = 120i + 310j + 60k

p = 29i + 18j + 21 k

Total cost = $10,320

Step-by-step explanation:

Let n and p represent the vectors for number of products and prices respectively.

Also, the coordinates i,j and k represent cashews, walnut and Brazil nut respectively.

The vector form of the total number of bags ordered and the cost are;

n = 120i + 310j + 60k

p = 29i + 18j + 21 k

We can obtain the total cost by obtaining the dot product of the two vectors.

Total cost = n.p = (120i + 310j + 60k).(29i + 18j + 21 k)

C = 120×29 + 310×18 + 60×21

C = $10,320

7 0

3 years ago

What is a radius f a circle with diameter of<br> 240mm

Yuki888 [10]

R=120 mm
diameter 240
so your answer is 120mm

6 0

3 years ago

Read 2 more answers

Sophie Ruth is eating a 50-gram chocolate bar which contains 30%, percent cocoa.

MrRissso [65]

Grams of cocoa =30% of 50g=(30/100) * 50 g=(30 * 50 g)/100=15 g

Soltution: 15 g

7 0

3 years ago

Read 2 more answers

Write the equation for a 3rd degree polynomial function that has roots at x = 1–3i and x = 2. The y‑intercept is (0,10). Write a

HACTEHA [7]

Answer:

$\displaystyle P(x)=-\frac{1}{2}(x^2+2x+10)(x-2)$

Step-by-step explanation:

<u>Factored Form Of Polynomials </u>

If we know the roots of a polynomial as

$\alpha _1,\alpha _2,\alpha _3$

the polynomial can be expressed in factored form as

$a(x-\alpha _1)(x-\alpha _2)(x-\alpha _3)$

We are given two of the three roots of the polynomial:

$\alpha _1=1-3i$

$\alpha _2=2$

The other root must be the conjugate of the complex root:

$\alpha _3=1+3i$

Recall the product of two complex conjugate numbers is

$(1-3i)(1+3i)=1^2-(3i)^2=1+9=10$

The required polynomial is

$P(x)=a\left[ x-(1-3i)\right ]\left[ x-(1+3i)\right ](x-2)$

$P(x)=a(x^2+2x+10)(x-2)$

This is the factored form of the polynomial where only real numbers appear

We need to find the value of a, such as

$P(0)=10$

$P(0)=a(0^2+2(0)+10)(0-2)=10$

$-20a=10$

Thus the value of a is

$\displaystyle a=-\frac{1}{2}$

The expression of the required polynomial is

$\boxed{P(x)=-\frac{1}{2}(x^2+2x+10)(x-2)}$

5 0

2 years ago