MATHEMATICS HIGH SCHOOL

(a) Solve 7( k - 3 ) = 3k - 5 (b) Expand and simplify (2x + 3 )( x - 8)

(c) Solve 7 - 3 f  = 2
                 4

Answers

Answer 1
Answer: a) 7(k-3)=3k-5\n 7k-21=3k-5\n 7k-3k=-5+21\n 4k=16\n k=\frac { 16 }{ 4 } \n k=4

b) (2x+3)(x-8)\n 2{ x }^( 2 )-16x+3x-24\n 2{ x }^( 2 )-13x-24

c) \frac { 7-3f }{ 4 } =2\n 7-3f=4\cdot 2\n 7-3f=8\n -3f=8-7\n -3f=1\n f=-\frac { 1 }{ 3 }
Answer 2
Answer: (a)\n7(k-3)=3k-5\n7(k)+7(-3)=3k-5\n7k-21=3k-5\ \ \ \ |add\ 21\ to\ both\ sides\n7k=3k+16\ \ \ \ |subtract\ 3k\ from\ both\ sides\n4k=16\ \ \ \ \ |divide\ both\ sides\ by\ 4\n\boxed{k=4}


(b)\n(2x+3)(x-8)=(2x)(x)+(2x)(-8)+(3)(x)+3(-8)\n\n=2x^2-16x+3x-24=\boxed{2x^2-13x-24}


(c)\n(7-3f)/(4)=2\ \ \ \ |multiply\ both\ sides\ by\ 4\n\n\not4^1\cdot(7-3f)/(\not4_1)=4\cdot2\n\n7-3f=8\ \ \ \ \ |subtract\ 7\ from\ both\ sides\n\n-3f=1\ \ \ \ \ |divide\ both\ sides\ by\ (-3)\n\n\boxed{f=-(1)/(3)}

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The cost to rent a construction crane is $750 per day plus $250 per hour of use. What is the maximum number of hours the crane can be used each day if the rental cost is not exceed $2500 per day ? Write amd solve the inequality

Answers

Let's use x to represent the number of hours since we don't know that.

750+250x<2500
Subtract 750 from both sides.
250x<1750
Divide both sides by 250.
x<7
The most hours that they can do is 7, or it can be less. 

Answer:

x = 7

Step-by-step explanation:

x= number of hours

"cost is not exceed $2500 per day"

not exceed = less than

750+250x < 2500

250x< 2500-750

250x < 1750

x < 1750/250

x<7

7 hours

Can you express 9/9 as a mixed number

Answers

9/9 simply equals 1 there is no mixed number

Find the product of (x + 3)2.

x2 + 6x + 9
x2 − 6x + 9
x2 − 9
x2 + 9

Answers

Hello,

(x+3)²=x²+2*3*x+3²=x²+6x+9

Answer A

Find the product of (x + 3)2.

Answer:

A) x2 + 6x + 9

(Took the exam)

The graph of y=15(x-92)^2+3 has the axis of symmetry at x=___

Answers

Maybe the answer is 2

Given f(x) = —x^2 - 13, find f(-6)

Answers

Answer:

f(-6) = -49

Step-by-step explanation:

f(-6) = -(-6)^2 - 13 = -(36) - 13 = -49.
f(-6)= -49



( f-(-6)= -(-6)^2-13= -(36)-13=-49 )

Use the Factor Theorem to determine whether the first polynomial is a factor of the second polynomial. x - 5; 3x2 + 7x + 40

Answers

Answer:

The polynomial (x -5) is not a factor of second polynomial 3x^2 + 7x + 40

Step-by-step explanation:

Factor theorem states that if you divide a polynomial p(x) by a factor x -a of that polynomial, then you will get a zero remainder.

i.,e p(x) = (x-a)q(x)   which means that if x - a is a factor of p(x), then the remainder, when we do synthetic division by  x= a, will be zero.

Determine whether the first polynomial is a factor of the second polynomial.

Given the polynomial:  f(x)=3x^2 + 7x + 40

For  x-5 to be a factor of f(x)=3x^2 + 7x + 40, the factor theorems implies that x = 5 must be a zero of f(x).

Now, to test whether x-5  is a factor;

Set x -5 = 0

⇒x = 5

Then,

we will use synthetic division method to divide f(x) by x =5

you can see the figure as shown below in the attachment.

Since, the remainder is 150 which is not equal to zero, then Factor theorem says that (x-5) is not a  factor of 3x^2 + 7x + 40

Answer:

No,(x-5) is not a factor of 3x^2+7x+40

Explanation:

Factor theorem states that if (x-a) is a factor of the function f(x) then f(a) = 0.

We can use this theorem to check whether a polynomial is a factor of other polynomial or not.

Further Explanation:

Here, we have to check if (x-5) is a factor of 3x^2+7x+40 or not.

For this we can use the above mentioned factor theorem.

In our case,

a = 5

and f(x)=3x^2+7x+40

So, we find f(5) and see if it is zero or not. If f(5) = 0 then (x-5) must be the factor the polynomial.

f(5)=[tex]3(5)^2+7(5)+40\n\n=75+35+40\n\n=150\neq0

Since, f(5) is not zero. Hence, from factor theorem, (x-5) is not a factor of 3x^2+7x+40

Learn More:

brainly.com/question/12482195 (Answered by Kudzordzifrancis)

brainly.com/question/11378552 (Answered by Alinakincsem)

Keywords:

Factor theorem, Remainder theorem.
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