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3 years ago

14

Write the ratio in simplest form in both fractions and in decimal form.

2 answers:

lara [203]3 years ago

8 0

Answer:

Fraction form for number 1: 1/4

Decimal form for number 1: 0.25

Fraction form for number 2: 1/16

Decimal form for number 2: 0.0625

Step-by-step explanation:

For the fraction form for number one, all we have to do is change 3:12 to 3/12. Then we just simplify. 1/4. The decimal for 1/4 is just .25, or one quarter.

For the second one, we need to change 4:64 into 4/64. Then we simplify to 1/16. The decimal is a little bit trickier but it comes out to 0.0625.

You're done!

Plz vote brainliest

ValentinkaMS [17]3 years ago

3 0

Answer:

Step-by-step explanation:

3 : 12

Both have table of 3 in common so

1 : 4 or 1/4

4 : 64

Both have table of 4 in common

1 : 16 or 1/16

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If a certain cannon is fired from a height of 8.8 meters above the ground, at a certain angle, the height of the cannonball ab

Dennis_Churaev [7]

Answer:

It would take approximately 6.50 second for the cannonball to strike the ground.

Step-by-step explanation:

Consider the provided function.

$h(t)=-4.9t^2+30.5t+8.8$

We need to find the time takes for the cannonball to strike the ground.

Substitute h(t) = 0 in above function.

$-4.9t^2+30.5t+8.8=0$

Multiply both sides by 10.

$-49t^2+305t+88=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are: $x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute a = -49, b = 305 and c=88

$t=\frac{-305+\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}=-\frac{-305+\sqrt{110273}}{98}\\t = \frac{-305-\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}= \frac{305+\sqrt{110273}}{98}$

Ignore the negative value of t as time can't be a negative number.

Thus,

$t=\frac{305+\sqrt{110273}}{98}\approx6.50$

Hence, it would take approximately 6.50 second for the cannonball to strike the ground.

6 0

3 years ago

+50 points quiz

Sedaia [141]

Answer:

wasting your points but hey these are the answers

Step-by-step explanation:

1) 72

2) 2

3) 30

4) 2

5 0

3 years ago

Read 2 more answers

I made a fort by two boxes. The first box is 4 meters long, 8 meters wide, and 8 meters high. The second box is 2 meters long, 7

ArbitrLikvidat [17]

Answer:

242 m³ of space is there in the fort.

Step-by-step explanation:

Given that,

The dimensions of first box is 4 meters long, 8 meters wide, and 8 meters high.

The dimensions of the second box is 2 meters long, 7 meters wide, and 1 meter high.

Space left = Volume of first box - volume of second box

= (4)(8)(8) - 2(7)(1)

= 242 m³

So, 242 m³ of space is there in the fort.

7 0

3 years ago

Simplify the expression:<br> 5q+–8q

vodka [1.7K]

Answer:

-3q

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Algebra I</u>

Terms

Step-by-step explanation:

<u>Step 1: Define</u>

5q + -8q

<u>Step 2: Simplify</u>

Rewrite: 5q - 8q
Combine like terms: -3q

3 0

3 years ago

Read 2 more answers

Find the general solution of the given differential equation.

Elina [12.6K]

Answer:

$y=-x\,cos\,x+Cx$

There are no transient terms.

Step-by-step explanation:

Given: $x\,\frac{dy}{dx} -y=x^2\,sin\,x$

To find: general solution of the differential equation and the transient terms in the general solution.

Solution:

For an equation of the form $\frac{dy}{dx}+yp(x)=q(x)$ ,

solution is given by $ye^{\int {p(x)} \, dx }$ = ∫ q(x) $e^{\int {p(x)} \, dx }$ dx

The given equation $x\,\frac{dy}{dx} -y=x^2\,sin\,x$ can be written as $\frac{dy}{dx}-\frac{y}{x}=x\,sin\,x$

Here,

$p(x)=\frac{-1}{x}\,,\,q(x)=x\,sin\,x$

$e^{\int{p(x)} \, dx } =e^{\int{\frac{-1}{x} } \, dx } =e^{-ln(x)} =e^{ln(x^{-1} )}=x^{-1}=\frac{1}{x}$

So,

the solution is $\frac{y}{x}=\int \frac{1}{x}x\,sin\,x\,dx$

$\frac{y}{x} =\int\,sin\,x\,dx\\\\\frac{y}{x} =-cos\,x+C\\y=-x\,cos\,x+Cx$

Here, C is a constant.

Transient term is a term such that it tends to 0 as x → ∞

Here, there does not exist any term that tends to 0 as x → ∞

So, there are no transient terms.

7 0

3 years ago