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Marysya12 [62]
2 years ago
12

0=-16t^2 +10 Solve for t

Mathematics
2 answers:
Veseljchak [2.6K]2 years ago
5 0

0=-16t^2 +10 \\\\16t^2=10\\\\t^2=\dfrac{10}{16}=\dfrac{5}{8}\\\\t=\sqrt{\dfrac{5}{8}} \vee t=-\sqrt{\dfrac{5}{8}}\\\\t=\dfrac{\sqrt5}{\sqrt8} \vee t=-\dfrac{\sqrt5}{\sqrt8}\\\\t=\dfrac{\sqrt5}{2\sqrt2} \vee t=-\dfrac{\sqrt5}{2\sqrt2}\\\\t=\dfrac{\sqrt{10}}{2\cdot2} \vee t=-\dfrac{\sqrt{10}}{2\cdot2}\\\\t=\dfrac{\sqrt{10}}{4} \vee t=-\dfrac{\sqrt{10}}{4}

KengaRu [80]2 years ago
4 0

Answer:

plus or minus √10 / 4

Step-by-step explanation:

0=-16t^2 +10 becomes easier to solve if we rewrite it as 16t^2 = 10.

Then t^2 = 10/16 = 5/8, and t = plus or minus √(5/8), or

                                             t = plus or minus (√5) / √8), or

                                             t = plus or minus (√5)√8         √5*2√2

                                                                         ------------- = -------------------

                                                                                 8                    8

or                                                                 plus or minus √10 / 4

                                           

You might be interested in
7g:1kg to its simplest form​
hammer [34]

Answer:

7kg:1kg=7:1 itself

Because 7 and 1 has one '1' as a common factor, so they can't be simplified

5 0
3 years ago
Form a quadratic polynomial whose zeroes are 3-√3/5 and 3+√3/5?​
barxatty [35]

Answer:

x^2 -6x + 222/25

Step-by-step explanation:

If the zeros are as above, then ;

x = 3-√3/5 or x = 3 + √3/5

Firstly, let’s represent √3/5 by b

Thus;

The two roots are ;

x = 3-b or x = 3 + b

so;

x+ b -3 and x -3-b

The quadratic equation is the product of the two

(x + b-3)(x - b -3)

x(x - b-3) + b(x -b -3) -3(x - b -3)

= x^2 -bx -3x + bx -b^2 -3b -3x + 3b + 9

Collect like terms and we are left with;

x^2 -6x -b^2 + 9

So let’s put back b = √3/5

x^2 -6x -(√3/5)^2 + 9

x^2 -6x -3/25 + 9

x^2 -6x + 222/25

3 0
3 years ago
At carl's combine diner, there are three size of coffee drinks regular (300ml), large (500ml) and extra large (800mL), and they
MariettaO [177]

Answer:

The number of regular, large, and extra-large drinks are 12, 15, and 10 respectively.

Step-by-step explanation:

Given that the cost for regular coffee drinks (300 ml)=$2.25

The cost for large coffee drinks (500 ml)=$3.25

The cost for extra large coffee drinks (800 ml)=$5.75

Let p,q, and r be the number of regular, large, and extra-large coffee sold.

As the diner sold a total of 37 coffees, so

p+q+r=37

r=37-p-q...(i)

The volume of p regular coffee = 300p ml

The volume of q  large coffee = 500q ml

The volume of r extra-large coffee = 800r ml

As the total volume of coffee sold was 19,100mi, so

300p+500q+800r=19100

By using equation (i)

300p+500q+800(37-p-q)=19100

300p+500q+800 x 37 - 800p - 800q=19100

-500p-300q=19100-29600

500p+300q=10500

500p=10500-300q

p=21-0.6q...(ii)

Now, the cost of p regular coffee=$2.25p

The cost of q large coffee=$3.25q

The cost of r extra-large coffee=$5.75r

As the amount of money made in coffee sales was $133.25, so

2.25p+3.25q+5.75r=133.25

By using equations (1)  we have

2.25p+3.25q+5.75(37-p-q)=133.25

2.25p+3.25q+212.75-5.75p-5.75q=133.25

3.50p+2.50q=79.5

From equation (ii)

3.5(21-0.6q)+2.50q=79.5

73.5-2.1q+2.5q=79.5

0.4q=79.5-73.5=6

q=6/0.4

q=15

From equation (ii)

p=21-0.6(15)

p=12

From equation (i)

r= 37-12-15

r=10

Hence, the number of regular, large, and extra-large drinks are 12, 15, and 10 respectively.

5 0
3 years ago
Please help i need help geometry is very hard for me i am dumb
stiks02 [169]

Answer:

G.  ABD = 74

H.  DBC = 206

I.  XYW = 33.75

J.  WYZ = 46.25

Step-by-step explanation:

For G and H: You have a straight line (ABC) with another line coming off of it, creating two angles (ABD and DBC).  A straight line has an angle of 180 degrees.  This means that the two angles from the straight line when combined will give you 180 degrees.  Solve for x.

ABD + DBC = ABC

(1/2x + 20) + (2x - 10) = 180

1/2x + 20 + 2x - 10 = 180

5/2x + 10 = 180

5/2x = 170

x = 108

Now that you have x, you can solve for each angle.

ABD = 1/2x + 20

ABD = 1/2(108) + 20

ABD = 54 + 20

ABD = 74

DBC = 2x - 10

DBC = 2(108) - 10

DBC = 216 - 10

DBC = 206

For I and J:  For these problems, you use the same concept as before.  You have a right angle (XYZ) that has within it two other angles (XYW and WYZ).  A right angle has 90 degrees. Combine the two unknown angles and set it equal to the right angle.  Solve for x.

XYW + WYZ = XYZ

(1 1/4x - 10) + (3/4x + 20) = 90

1 1/4x - 10 + 3/4x + 20 = 90

2x + 20 = 90

2x = 70

x = 35

Plug x into the angle values and solve.

XYW = 1 1/4x - 10

XYW = 1 1/4(35) - 10

XYW = 43.75 - 10

XYW = 33.75

WYZ = 3/4x + 20

WYZ = 3/4(35) + 20

WYZ = 26.25 + 20

WYZ = 46.25

3 0
3 years ago
Is y=7x-3 have a proportional relationship
Black_prince [1.1K]

Answer:

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Step-by-step explanation:

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8 0
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