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3 years ago

T+2/3t-1/3 I really need this question to be answered

Mathematics

1 answer:

Murrr4er [49]3 years ago

5 0

Equation:

x(x - 5) + 3(x + 5)

Steps:

x(x - 5) + 3(x + 5)

Expand:

x(x - 5 ): x^2 - 5x

x^2 - 5x + 3(x + 5)

Expand:

3(x + 5): 3x + 15

x^2 - 5x + 3x + 15

Add Similar Elements:

-5x + 3x

= -2x

Answer x^2 - 2x + 15 Doesn't Factor

Hope that helps!!! : )

You might be interested in

I now have another question to ask.

marissa [1.9K]

Answer:

23.6 ft

Step-by-step explanation:

Sketch a right triangle representing this situation. The length of the hypotenuse is 26 ft and the angle of elevation from ground to top of ladder is 65°. The "opposite side" is the reach of the ladder, which we'll call x.

Then:

opp

sin 65° = ----------
26 ft

or (26 ft)(sin 65°) = opp side = height off the ground of top of ladder.

Evaluating this, we get:

(26 ft)(0.906) = 23.56 ft, or, rounded off, 23.6 ft

The ladder reaches 23.6 ft up the side of the building.

8 0

3 years ago

Scott is on his school's academic team. On average, it takes Scott 4 minutes, with a standard deviation of 0.25 minutes, to solv

uysha [10]

Answer:

15.87% is the chance that Scott takes more than 4.25 minutes to solve a problem at an academic bowl.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4 minutes

Standard Deviation, σ = 0.25 minutes

We standardize the given data.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(more than 4.25 minutes to solve a problem)

$P( x > 4.25) = P( z > \displaystyle\frac{4.25 - 4}{0.25}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,

$P(x > 4.25) = 1 - 0.8413 = 0.1587 = 15.87\%$

Thus,15.87% is the chance that Scott takes more than 4.25 minutes to solve a problem at an academic bowl.

8 0

2 years ago

I need help pls help me

agasfer [191]

Answer:

Step-by-step explanation:

2ft=24in

The first day 24(1/6)=4in fell

The second day 24(7/12)=14in fell

The third day 24-4-14=6in fell

6 0

3 years ago

Read 2 more answers

A 30 gram sample of a substance that's used to sterilize surgical instruments has a k-value of 0.1235. Find the substances half

jok3333 [9.3K]

Answer : The substances half life in days is, 5.6 days.

Step-by-step explanation :

Half life : It is defined as the amount of time taken by a radioactive material to decay to half of its original value.

All radioactive decays follow first order kinetics.

The relation between the half-life and rate constant is:

$k=\frac{0.693}{t_{1/2}}$

where,

k = rate constant = 0.1235 per days

$t_{1/2}$ = half-life

Now put all the given values in the above formula, we get:

$0.1235\text {days}^{-1}=\frac{0.693}{t_{1/2}}$

$t_{1/2}=5.6\text{ days}$

Thus, the substances half life in days is, 5.6 days.

5 0

3 years ago

Better Products, Inc., manufactures three products on two machines. In a typical week, 40 hours are available on each machine. T

Kaylis [27]

Answer:

z (max) = 1250 $

x₁ = 25 x₂ = 0 x₃ = 25

Step-by-step explanation:

Profit $ mach. 1 mach. 2

Product 1 ( x₁ ) 30 0.5 1

Product 2 ( x₂ ) 50 2 1

Product 3 ( x₃ ) 20 0.75 0.5

Machinne 1 require 2 operators

Machine 2 require 1 operator

Amaximum of 100 hours of labor available

Then Objective Function:

z = 30*x₁ + 50*x₂ + 20*x₃ to maximize

Constraints:

1.-Machine 1 hours available 40

In machine 1 L-H we will need

0.5*x₁ + 2*x₂ + 0.75*x₃ ≤ 40

2.-Machine 2 hours available 40

1*x₁ + 1*x₂ + 0.5*x₃ ≤ 40

3.-Labor-hours available 100

Machine 1 2*( 0.5*x₁ + 2*x₂ + 0.75*x₃ )

Machine 2 x₁ + x₂ + 0.5*x₃

Total labor-hours :

2*x₁ + 5*x₂ + 2*x₃ ≤ 100

4.- Production requirement:

x₁ ≤ 0.5 *( x₁ + x₂ + x₃ ) or 0.5*x₁ - 0.5*x₂ - 0.5*x₃ ≤ 0

5.-Production requirement:

x₃ ≥ 0,2 * ( x₁ + x₂ + x₃ ) or -0.2*x₁ - 0.2*x₂ + 0.8*x₃ ≥ 0

General constraints:

x₁ ≥ 0 x₂ ≥ 0 x₃ ≥ 0 all integers

The model is:

z = 30*x₁ + 50*x₂ + 20*x₃ to maximize

Subject to:

0.5*x₁ + 2*x₂ + 0.75*x₃ ≤ 40

1*x₁ + 1*x₂ + 0.5*x₃ ≤ 40

2*x₁ + 5*x₂ + 2*x₃ ≤ 100

0.5*x₁ - 0.5*x₂ - 0.5*x₃ ≤ 0

-0.2*x₁ - 0.2*x₂ + 0.8*x₃ ≥ 0

x₁ ≥ 0 x₂ ≥ 0 x₃ ≥ 0 all integers

After 6 iterations with the help of the on-line solver AtomZmaths we find

z (max) = 1250 $

x₁ = 25 x₂ = 0 x₃ = 25

6 0

3 years ago