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4 years ago

What is three times a number b and 33 is at least 35

Mathematics

1 answer:

cestrela7 [59]4 years ago

8 0

The answer is b=2/3. You subtract 33 from 35. then divide 2 by 3.

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Let the function be f defined by f (x) = 12 - 5x

marissa [1.9K]

(a) No

$f(x) = 12 - 5x$

$f(0) = 12 - 5(0)$

$f(x) = 12 - 0$

$f(x) = 12$

Hence,

f(0) is unequal to 0.

(b)

$f(x) = 12 - 5x$

$f( - 1) = 12 - 5( - 1)$

$f( - 1) = 12 + 5$

$f( - 1) = 17$

(c)

$f(x) = 12 - 5x$

$f(x) = 6$

$6 = 12 - 5x$

$6 - 12 = - 5x$

$- 6 = - 5x$

$6 = 5x$

$\frac{6}{5} = x$

7 0

3 years ago

How would i solve these two problems?

Montano1993 [528]

Number 17 9216 because you take 96*4

8 0

4 years ago

Share 120 in the ratio 1:5

zhuklara [117]

What are you asking here i think it should be 20:100

5 0

3 years ago

What is the exact value of sin -840

dsp73

First subtract 2 revolutions from the - 840 :- (- (840-720) = -120

This gives sin -120 which is in the 3rd quadrant of the unit circle

sin -120 = - sin 60 = - sqrt3/2

7 0

4 years ago

Identify each function that has a remainder of -3 when divided x+6

Sergeu [11.5K]

Answer:

Step-by-step explanation:

According to remainder theorem, you can know the remainder of these polynomials if you plug in x = -6 into them.

So we will plug in -6 into x of all the polynomials ( A through D) and see which one equals -3.

For A:

$x^5 + 2x^2 - 30x + 30\\=(-6)^5 + 2(-6)^2 - 30(-6) + 30\\=-7494$

For B:

$x^4 + 4x^3 - 21x^2 - 53x + 12\\=(-6)^4 + 4(-6)^3 - 21(-6)^2 - 53(-6) + 12\\=6$

For C:

$x^3 - 10x^2 - 7\\=(-6)^3 - 10(-6)^2 - 7\\=-583$

For D:

$x^4 + 6x^3 - 10x - 63\\=(-6)^4 + 6(-6)^3 - 10(-6) - 63\\=-3$

The only function that has a remainder of -3 when divided by x + 6 is the fourth one, answer choice D.

5 0

3 years ago