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3 years ago

What is 31/4 + -11/2 + 21/4

Mathematics

2 answers:

Basile [38]3 years ago

6 0

31/4 + -11/2 + 21/4 = 7 1/2
Hope this helps!

Andrei [34K]3 years ago

6 0

Hope this helped, and GOOD LUCK! :D

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When a ball is thrown into the air, the path that the ball travels is modeled by the parabola y − 5 = −0.05(x − 10)2, measured i

beks73 [17]

Answer:

the maximum height in feet the ball reaches 5ft

Step-by-step explanation:

4 0

3 years ago

A light bulb consumes 1440 watt hours per day. How many watt hours does it consume in 5 days and 18 hours

Maurinko [17]

Answer:

8,280 watt hours

Step-by-step explanation:

1440/24 = 60 watt hours per hour

5*24=120 hours in 5 days

120+18=138

138*60=8,280 watt hours

4 0

3 years ago

Is (6,-10) a solution of y=3x-8

Vitek1552 [10]

You can easily test this if you know that (6, -10) corresponds to (X, Y). Knowing this, you can:

X = 6
Y = -10

you put this into your equation:

-10 = 3*6 - 8

calculate it:

-10 = 18 - 8
-10 = 10
This is not true of course, -10 is not equal to 10. Therefore, (6, -10) is not a solution of y = 3x-8 :)

6 0

3 years ago

Annita, Gary, and Tara all run races. Annita has 7 less than 4 times the number of race medals as Tara. Gary has 13 more than 2

harkovskaia [24]

Answer:

4 t - 7 = 2 t+ 13 is the required equation.

The number of race medals Tara has is t = 20 medals.

Step-by-step explanation:

Let the number of medals Tara has = t medals

So, the number of medal Anita has = 4( Medals of Tara) - 7

= 4t - 7

And the number of medals Gary has = 2 times (Medals of Tara) + 13

= 2(t) + 13 = 2t + 13

Now, Annita and Gary has same number of medals.

⇒ 4t - 7 = 2t+ 13

or, 4t - 2t = 7 + 13

⇒ 2t = 20

⇒ t = 20/2 = 10

or t = 10

Hence, the number of race medals Tara has is t = 20 medals

6 0

3 years ago

Find the least common multiple (LCM) of 8y^6+ 144y^5+ 640y^4 and 2y^4 + 40y^3 + 200y^2.

Mice21 [21]

Answer:

LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

Step-by-step explanation:

Making factors of $8y^{6}+ 144y^{5}+ 640y^{4}$

Taking $8y^{4}$ common:

$\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)$

Using factorization method:

$\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times 4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)$

Now, Making factors of $2y^{4} + 40y^{3} + 200y^{2}$

Taking $2y^{2}$ common:

$\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)$

Using factorization method:

$\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10) ............ (2)$

The underlined parts show the Highest Common Factor(HCF).

i.e. HCF is $2y^{2} (y+ 10)$ .

We know the relation between LCM, HCF of the two numbers 'p' , 'q' and the numbers themselves as:

$HCF \times LCM = p \times q$

Using equations (1) and (2): $\Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)$

Hence, LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

5 0

3 years ago