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3 years ago

1.4t - 0.4 (t- 3.1) =5.8

1 answer:

7 0

1 Expand
1.4t-0.4t+1.24=5.8

2 Simplify 1.4t-0.4t+1.241.4t−0.4t+1.24 to t+1.24t+1.24
t+1.24=5.8

3 Subtract 1.241.24 from both sides
t=5.8−1.24

4 Simplify 5.8-1.245.8−1.24 to 4.564.56
t=4.56

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I need help with this work

STatiana [176]

Answer:

G. 245.04cm²

Step-by-step explanation:

The shortest way to solve this problem would be:

total surface area = 2 * π * r² + (2 * π * r) * h

total surface area = 2 * π * r * (r + h).

Diameter = 6

Radius = 3

Height = 10

The radius is half the diameter

TSA = Total surface area

TSA = 2 π 3(3+10)

TSA = (2)(3.141593)(3)(3+10)

TSA = (6.283185)(3)(3+10)

TSA = 18.849556(3+10)

TSA = (18.849556)(13)

TSA = 245.044227

TSA = 245.04

Or you could do the longer way, which would be:

Step 1: Find the surface area of the curved part of the cylinder

Surface Area (Curved) = 2 x π x Radius x Height

Step 2: Find the surface area of the two ends of the cylinder

Surface Area (Ends) = 2 x π x Radius²

Step 3: Add the surface are of the curved part to the surface area of the ends

Surface Area (Total) = Surface Area (Curved) + Surface Area (Ends)

For the cylinder in the picture:

A cylinder has a height of 10 and a diameter of 6

Hence, Radius = d/2 = 6/2 = 3 cm

Surface Area (Curved) = 2 x pi x 3 x 10

Surface Area (Curved) = 188.495559

Surface Area (Ends) = 2 x pi x 3²

Surface Area (Ends) = 2 x pi x 25

Surface Area (Ends) = 56.548668

Surface Area (Total) = 245.044227

245.04 is the result of rounding 245.044227 to the nearest hundredth

7 0

2 years ago

−9x (5x+4 ) +10 (x+3) can someone please solve his for me thank u step by step please i really need to find out how to do it

Ivahew [28]

Answer:

-45 $x^{2}$ -26x+30

Step-by-step explanation:

So first you let's divide this equation into three parts.

Part one -9x(5x+4)

Step one: you have to distribute the -9x to the numbers in the parentheses. That would leave us with a -9x*5x = -45 $x^{2}$ and -9x*4 = -36x
Step two: Put the answers together. That would leave us with -45 $x^{2}$ -36x.

Part two 10(x+3)

Step one: you have to distribute the 10 to the numbers in the parentheses. That would leave us with a 10*x = 10x and 10*3 = 30
Step two: Put the answers together. That would leave us with 10x+30

Part three solve

Step one: combine and put together what you have solved for. That would leave us with a -45 $x^{2}$ -36x+ 10x+30
Step two: combine like terms. The like terms here are -36x and 10x. When you combine them you would get -26x.
Step three: Write the equation in standard form. Therefore the answer is -45 $x^{2}$ -26x+30.

Let me know if anything is confusing!

5 0

2 years ago

Answer ,how to do it.

guapka [62]

Answer:

The system has two solutions:

x₁ = 5 ⇒ y₁ = -10

x₂ = -2 ⇒ y₂ = 11

Step-by-step explanation:

Let's solve the system of equations, this way:

y = -3x + 5

y = x ² - 6x - 5

Replacing y in the 2nd equation:

y = x ² - 6x - 5

-3x + 5 = x ² - 6x - 5

x ² - 3x - 10 = 0

Solving for x, using the quadratic formula:

(3 +/- √(9 -4 * 1 * -10))/2 * 1

(3 +/- √9 + 40)/2

(3 +/- √49)/2

(3 +/- 7)/2

x₁ = 10/2 = 5

x₂ = -4/2 = -2

x₁ = 5 ⇒ y₁ = -10

x₂ = -2 ⇒ y₂ = 11

As we can see the system has two different solutions

8 0

3 years ago

Suppose you borrow money for 6 months. If the interest rate is compounded monthly, the formula

Alinara [238K]

Answer:

$1045.85.

Step-by-step explanation:

A=p(1+r/12)^6

Here r = 0.09, P = $1000 so

Amount of money needed to pay off the loan =

1000(1 + 0.09/12)^6

= $1045.85.

6 0

3 years ago

The function below models the voltage, in volts, of a certain alternating current after x seconds, where A and b are positive co

Katarina [22]

First of all, we can just ignore A, it has no effect but to vertically stretch our cosine.

If it was only $f(x)=cosx$ , the function would be invertible as long as it's confined between $0$ and $\pi$ . Now, the argument of our cosine is not $x$ but $bx$ . It means that it won't stop at $\pi$ , but at $\frac{\pi}{b}$ .

Another way to think about it, "what should i replace x with so I get $\pi$ inside the cosine?"

4 0

3 years ago