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2 years ago

What is the leading coefficient of the function f(x)=3x^5+6x^4-x-3

1 answer:

3 0

A coefficient is a number that is multiplied by a variable (a letter). And " leading " means the first one..

f(x) = 3x^5 + 6x^4 - x - 3......so ur leading coefficient is going to be 3

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Y = 7x + 5 y = 7x - 3 These 2 lines are ? O parallel Operpendicular Oneither

amm1812

Answer:

Parallel

Step-by-step explanation:

These 2 lines are parallel because they have the same slope (7).

8 0

3 years ago

Divide using synthetic division: (2x^4+2x^3+-12x^2+x+6)/(x+3)

kap26 [50]

Synthetic division yields

..... || 2 2 -12 1 6

-3 || -6 12 0 -3

= = = = = = = = = = = = = =

..... || 2 -4 0 1 3

which translates to

$\dfrac{2x^4+2x^3-12x^2+x+6}{x+3}=2x^3-4x^2+1+\dfrac3{x+3}$

5 0

2 years ago

A retailer returned 2700 defective items to the manufacturer and received a credit for the retail price of $0.83 or 834/34 per i

Helga [31]

Answer:

credit gain by retailer is $2257.39

Step-by-step explanation:

Given data:

Retaile price 83 4/3 c or 84.33c

Let Discount rate 6/7

Discount amount = (6/7)% of 84.33 c = 0.7228 C

Net credit received = 84.33c - 0.7228c = 83.607c

Net credit received = 83.607 c

Item required = 2700

Total discount = 2700 × 83.0607 = 225739.44c

Total discount = $2257.39

credit received by retailer is $2257.39

8 0

2 years ago

What expressions are equivalent to 3(n+6)

KengaRu [80]

3n+3*6 which simplifies to 3n+18

8 0

3 years ago

Write the expression in terms of sine and cosine, and then simplify so that no quotients appear in the final expression. sec bet

Ede4ka [16]

Answer:

Step-by-step explanation:

The expression given is: sec β - cos β ( i will write "beta" instead of the symbol β)

We know that $sec(x)=\frac{1}{cos(x)}$

Substituting this into the the expression and doing some algebra manipulation, we have:

$sec(beta)-cos(beta)\\=\frac{1}{cos(beta)}-cos(beta)\\=\frac{1-cos^{2}(beta)}{cos(beta)}$

Using the identity $cos^{2}(x)+sin^{2}(x)=1\\$ (also $sin^{2}(x)=1-cos^{2}(x)$ ), we can now write:

$\frac{1-cos^{2}(beta)}{cos(beta)}\\=\frac{sin^2(beta)}{cos(beta)}\\=\frac{sin(beta)*sin(beta)}{cos(beta)}\\=\frac{sin(beta)}{cos(beta)}*sin(beta)$

We know $tan(x)=\frac{sin(x)}{cos(x)}$ , substituting this, we have:

$\frac{sin(beta)}{cos(beta)}*sin(beta)\\=tan(beta)*sin(beta)$

Answer choice E is the right answer.

3 0

2 years ago