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3 years ago

(15)Simplify the expression below. Show your work -(-7y+12)

2 answers:

6 0

15. -(-7y + 12) = 7y - 12

16. 1/a = 16/18
cross multiply
16a = 18
a = 18/16 = 9/8

17. 8x - 12 = 4x + 24
8x - 4x = 24 + 12
4x = 36
x = 36/4
x = 9

18. -6b > 42 4b > -4
b < 42/-6 b > -4/4
b < - 7 b > -1
so b < -7 and b > -1

19. 6 more then the product of 8 and n
6 + 8n

20. 45 = 3b + 69
45 - 69 = 3b
-24 = 3b
-24/3 = b
-8 = b

Rudik [331]3 years ago

5 0

(15) -(-7y + 12) = +7y - 12 (17) 8x - 12 = 4x + 24---> 8x - 4x = 24 + 12 ---> 4x = 36 ---> x = 9 (18) -6b > 42 ---> b<42/-6 ---> b<8 & 4b > 44 --> b> 11. So, 8>b>11 (20) 45 = 3b + 69 --> 3b = -24 ---> b = -8. Not sure what to do for (16) & (19)

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Q = 3.
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3 years ago

In triangle abc shown below side ab is 6 and side ac is 4

kaheart [24]

Question is Incomplete, Complete question is given below:

In Triangle ABC shown below, side AB is 6 and side AC is 4.

Which statement is needed to prove that segment DE is parallel to segment BC and half its length?

Answer

Segment AD is 3 and segment AE is 2.

Segment AD is 3 and segment AE is 4.

Segment AD is 12 and segment AE is 4.

Segment AD is 12 and segment AE is 8.

Answer:

Segment AD is 3 and segment AE is 2.

Step-by-step explanation:

Given:

side AB = 6

side AC = 4

Now we need to prove that segment DE is parallel to segment BC and half its length.

Solution:

Now AD + DB = AB also AE + EC = AC

DB = AB - AD also EC = AC - AE

Now we take first option Segment AD is 3 and segment AE is 2.

Substituting we get;

DB = 6-3 = 3 also EC = 4-2 =2

From above we can say that;

AD = DB and EC = AE

So we can say that segment DE bisects Segment AB and AC equally.

Hence From Midpoint theorem which states that;

"The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is congruent to one half of the third side."

Hence Proved.

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3 years ago

3. The curve C with equation y=f(x) is such that, dy/dx = 3x^2 + 4x +k

Andreas93 [3]

a. Given that y = f(x) and f(0) = -2, by the fundamental theorem of calculus we have

$\displaystyle \frac{dy}{dx} = 3x^2 + 4x + k \implies y = f(0) + \int_0^x (3t^2+4t+k) \, dt$

Evaluate the integral to solve for y :

$\displaystyle y = -2 + \int_0^x (3t^2+4t+k) \, dt$

$\displaystyle y = -2 + (t^3+2t^2+kt)\bigg|_0^x$

$\displaystyle y = x^3+2x^2+kx - 2$

Use the other known value, f(2) = 18, to solve for k :

$18 = 2^3 + 2\times2^2+2k - 2 \implies \boxed{k = 2}$

Then the curve C has equation

$\boxed{y = x^3 + 2x^2 + 2x - 2}$

b. Any tangent to the curve C at a point (a, f(a)) has slope equal to the derivative of y at that point:

$\dfrac{dy}{dx}\bigg|_{x=a} = 3a^2 + 4a + 2$

The slope of the given tangent line $y=x-2$ is 1. Solve for a :

$3a^2 + 4a + 2 = 1 \implies 3a^2 + 4a + 1 = (3a+1)(a+1)=0 \implies a = -\dfrac13 \text{ or }a = -1$

so we know there exists a tangent to C with slope 1. When x = -1/3, we have y = f(-1/3) = -67/27; when x = -1, we have y = f(-1) = -3. This means the tangent line must meet C at either (-1/3, -67/27) or (-1, -3).

Decide which of these points is correct:

$x - 2 = x^3 + 2x^2 + 2x - 2 \implies x^3 + 2x^2 + x = x(x+1)^2=0 \implies x=0 \text{ or } x = -1$

So, the point of contact between the tangent line and C is (-1, -3).

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2 years ago

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Romashka-Z-Leto [24]

Answer:

i think {x,y} = {-5,-2}

Step-by-step explanation:

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2 years ago

Read 2 more answers

The sum of four and the product of three and a number x.

KIM [24]

Answer:

Step-by-step explanation:

The product of three and a number x: 3 · x = 3x

The sum of four and the product of three and a number x:

4 + 3x

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3 years ago