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3241004551 [841]
3 years ago
6

Simplify -3c+5d+3c-d

Mathematics
2 answers:
Rina8888 [55]3 years ago
8 0

Answer:

4d

-3x+3c+5d-d

=0+4d

=4d

EastWind [94]3 years ago
4 0

Answer:

4d

Step-by-step explanation:

-3c+5d+3c-d

=-3c+3c+5d-d

=4d

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Grandma uses 2/3 of a pumpkin to make a pumpkin pie. How many pumpkins should she buy to make 4 pumpkin pies?​
Pachacha [2.7K]

she needs to use 8/12 of pumpkins

4 0
3 years ago
QUESTION: Determine if the statement is true or false? if false give an counterexample.
Jobisdone [24]

Answer:

  • False

Step-by-step explanation:

<u>Actual question is:</u>

  • logf 4(3d) + log4 1 = log4 3d

<u>Solution:</u>

  • logf 4 + logf (3d)  + 0 = log4 (3d)
  • logf 4 + logf (3d) = log4 (3d)

<u>If we assume f = 4, then</u>

  • log4 4 + log4 (3d) = log4 (3d)
  • log4 4 = 0

but log4 4 = 1

  • 1 = 0 is impossible

Therefore the statement is False

5 0
2 years ago
3(2x + 11) and (3x + 15)(2)
algol [13]

Answer:

3(2x+11)= 6x+33  and (3x+15)(2) = 6x+30

Step-by-step explanation:

Use distributive property to multiply the outside factor to each factor inside the parenthesis.  

3(2x+11)

(3*2x)+(3*11)

6x+33

3 0
3 years ago
Write the equation of the line in slope-intercept form.
Natasha_Volkova [10]

Answer:

y=-1/2x-2

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
An object is propelled upward from the top of a 300 foot building. The path that the object takes as it falls to the ground can
serg [7]

Answer:

As per the statement:

The path that the object takes as it falls to the ground can be modeled by:

h =-16t^2 + 80t + 300

where

h is the height of the objects and

t is the time (in seconds)

At t = 0 , h = 300 ft

When the objects hit the ground, h = 0

then;

-16t^2+80t+300=0

For a quadratic equation: ax^2+bx+c=0         ......[1]

the solution for the equation is given by:

x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}

On comparing the given equation with [1] we have;  

a = -16 ,b = 80 and c = 300

then;

t= \frac{-80\pm \sqrt{(80)^2-4(-16)(300)}}{2(-16)}

t= \frac{-80\pm \sqrt{6400+19200}}{-32}

t= \frac{-80\pm \sqrt{25600}}{-32}  

Simplify:

t = -\frac{5}{2} = -2.5 sec and t = \frac{15}{2} = 7.5 sec

Time can't be in negative;

therefore, the time it took the object to hit the ground is 7.5 sec

8 0
3 years ago
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