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fredd [130]
3 years ago
8

A. -4 <= x <= 0 b. y >= -3 c. all real numbers 4. {-4,0}

Mathematics
1 answer:
zimovet [89]3 years ago
8 0

Answer:

-4 <= x <= 0

Step-by-step explanation:

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Find the values for a and b that would make the following equation true.
ale4655 [162]

Answer:

\left(ax^2\right)\left(-6x^b\right)=12x^5\\\\(-6a)x^{2+b}=12x^5\to -6a=12\ and\ 2+b=5\\\\-6a=12\ \ \ |:(-6)\\a=-2\\\\2+b=5\ \ \ |-2\\b=3

Answer:\ a=-2;\ b=3

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Simplify (4x+5y)(2x-3y)+3xy<br><br> Help
amid [387]

Answer:

8x² - 15y² + xy

Step-by-step explanation:

(4x + 5y) (2x - 3y) + 3xy

multiplying the terms in brackets

(4x) (2x - 3y) + (5y) (2x - 3y) + 3 xy

multiplying with each terms inside the bracket

(4x)(2x) - (4x) (3y) + (5y) (2x) - (5y) (3y) + 3xy

doing the product each of the pair of terms

8x² - 12xy + 10xy - 15y² + 3xy

taking the sum of terms with coefficient "xy"

8x² - 15y² -2xy + 3xy

8x² - 15y² + xy


3 0
3 years ago
How would you describe the relationship between miles driven per week and car age, as shown in the scatterplot below? Choose eve
QveST [7]

Answer:

Negative and nonlinear

Step-by-step explanation:

hope this helped

5 0
2 years ago
I’m confused and don’t have an angle tool to even try to do the question
yanalaym [24]
<h3>Answer:   40</h3>

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

Explanation:

JQ is longer than QN. We can see this visually, but the rule for something like this is the segment from the vertex to the centroid is longer compared to the segment that spans from the centroid to the midpoint.

See the diagram below.

The ratio of these two lengths is 2:1, meaning that JQ is twice as long compared to QN. This is one property of the segments that form when we construct the centroid (recall that the centroid is the intersection of the medians)

We know that JN = 60

Let x = JQ and y = QN

The ratio of x to y is x/y and this is 2/1

x/y = 2/1

1*x = y*2

x = 2y

Now use the segment addition postulate

JQ + QN = JN

x + y = 60

2y + y = 60

3y = 60

y = 60/3

y = 20

QN = 20

JQ = 2*y = 2*QN = 2*20 = 40

--------------

We have

JQ = 40 and QN = 20

We see that JQ is twice as larger as QN and that JQ + QN is equal to 60.

7 0
3 years ago
A boat on the ocean is 2 mi from the nearest point on a straight? shoreline; that point is 15 mi from a restaurant on the shore.
Sladkaya [172]

Answer:

a) x  =  0,70 miles

T  =  1/2* (√4 + x²)  +( 15 - x )/3   Objective Function to minimize

b) V(min)  = 2.59 m/hr

Step-by-step explanation:

Let assume the boat is in Point A,  she lands in point B, and R  is the restaurant

Let call  x distance between the point O in  which perpendicular line from the boat get to shoreline, and the point were she land.

We know  distance is

d = v*t       ⇒  t = d/v

She rows at 2 miles/hr  and walk at 3 miles /hr

According to that she takes

c (hypotenuse) of right triangle  AOB/ 2 rowing

and  15 - x /3  walking

Total time is:

T  =  c/2  + ( 15-x )/3          c =√(2)² + x²     ⇒  T = [√(2)² + x²  ] /2  +  ( 15-x )/3  

T  =  1/2* (√4 + x²)  +( 15 - x )/3   (1)

And that is  the Objective function to minimize

a) Taking derivatives on both sides of the equation we get

T´(t)  =  x /( √4 + x²)  - 1/3    ⇒   T´(t)  = 0     x /( √4 + x²)  - 1/3 = 0

3*x - √( √4 + x²) = 0     ⇒  3*x  =  √( √4 + x²)

Squaring both sides

9x²  =  4 + x²     ⇒  8x²  = 4       x² = 1/2      x  =  0,70 miles

If we plug this value in the Objective function we will get the minimum time

1/2* (√4 + x²)  +( 15 - x )/3    ⇒ [√ 4  + 0,5] /2 +  14,3/3 = T (min)

T(min)  =  1.06 + 4.77  = 5.83 hr

Distance L (hypotenuse of right triangle AOR)

L = √(2)²  + (15)²   L  = 15,13 miles

And that distance have to be traveled at least in 5.83 hr rowing

Then  as  v = d/t      V(min)  =  15.13/ 5.83   ⇒    V(min)  = 2.59 m/hr

6 0
3 years ago
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