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8090 [49]
4 years ago
8

All the names that apply to the number -1.5

Mathematics
1 answer:
Blababa [14]4 years ago
8 0
-3/2 -15/10 this is called a negative number. Negative 1 and 5 tenths
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Solve the equatuon<br><br> x/2-(2/(x+1))=1<br><br> x=?
Snezhnost [94]

\frac{x}{2}-\frac{2}{x+1}=1

We have 2 denominators that we need to get rid of. Whenever there are the denominators, all we have to do is multiply all whole equation with the denominators.

Our denominators are both 2 and x+1. Therefore, we multiply the whole equation by 2(x+1)

\frac{x}{2}[2(x+1)]-\frac{2}{x+1}[2(x+1)] = 1[2(x+1)]

Then shorten the fractions.

\frac{x}{2}[2(x+1)]-\frac{2}{x+1}[2(x+1)] = 1[2(x+1)]\\x(x+1)-2(2)=1(2x+2)

Distribute in all.

x^2+x-4=2x+2

We should get like this. Because the polynomial is 2-degree, I'd suggest you to move all terms to one place. Therefore, moving 2x+2 to another side and subtract.

x^2+x-4-2x-2=0\\x^2-x-6=0\\

We are almost there. All we have to do is, solving for x by factoring. (Although there are more than just factoring but factoring this polynomial is faster.)

(x-3)(x+2)=0\\x=3,-2

Thus, the answer is x = 3, -2

7 0
3 years ago
It’s not 3 but still need help with it
Leto [7]

Answer:

13

Step-by-step explanation:

5-(-8)=13

6 0
3 years ago
Read 2 more answers
Please help 6th grade math i will give brainliest to answer with worked shown
Wewaii [24]

Answer:

64

Step-by-step explanation:

She drinks 4 ounces every 30 minutes.

4*2= 8 She drinks 8 ounces in an hour.

8*8=64 She drinks 64 ounces in 8 hours

8 0
3 years ago
Doug hits a baseball straight towards a 15 ft high fence that is 400 ft from home plate. The ball is hit 2.5ft above the ground
Scorpion4ik [409]

Answer:

125.4\ \text{m/s}

Step-by-step explanation:

u = Initial velocity of baseball

\theta = Angle of hit = 30^{\circ}

x = Displacement in x direction = 400 ft

y = Displacement in y direction = 15 ft

y_0 = Height of hit = 2.5 ft

a_y = g = Acceleration due to gravity = 32.2\ \text{ft/s}^2

t = Time taken

Displacement in x direction

x=u_xt\\\Rightarrow x=u\cos\theta t\\\Rightarrow t=\dfrac{x}{u\cos\theta}\\\Rightarrow t=\dfrac{400}{u\cos30^{\circ}}\\\Rightarrow t=\dfrac{400}{u\dfrac{\sqrt{3}}{2}}\\\Rightarrow t=\dfrac{800}{u\sqrt{3}}

Displacement in y direction

y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow y=y_0+u\sin\theta t+\dfrac{1}{2}a_yt^2\\\Rightarrow 15=2.5+u\sin30^{\circ}(\dfrac{800}{u\sqrt{3}})+\dfrac{1}{2}\times -32.2\times (\dfrac{800}{u\sqrt{3}})^2\\\Rightarrow \dfrac{400}{\sqrt{3}}-\dfrac{10304000}{3u^2}-12.5=0\\\Rightarrow 218.44=\dfrac{10304000}{3u^2}\\\Rightarrow u=\sqrt{\dfrac{10304000}{3\times218.44}}\\\Rightarrow u=125.4\ \text{m/s}

The minimum initial velocity needed for the ball to clear the fence is 125.4\ \text{m/s}

5 0
3 years ago
Assume that the weight loss for the first month of a diet program varies between 6 pounds and 14 pounds, and is spread evenly ov
Llana [10]

Hello!

For this uniform distribution, the shape of the density function is a line. We can begin by identifying key elements of the distribution:

Range: 6 to 14 pounds

Formula of density function:

y = 1/Range

1 - (14 - 6) = 1/8 = 0.125

f(x) = 0.125

1)

P(x = 7)

A uniform distribution is a continuous distribution, so the probability of x being an EXACT value is approximately 0.

<u>This is the same as if we took the following integral. (Finding the area under a curve with the same start and end point)</u>
\int\limits^a_a {f(x)} \, dx  = 0

Therefore, P(x = 7) = 0.

2)
We can think of this as finding the area of a rectangle.

The width would be the difference between 8.25 pounds and 12 pounds:
W = 12 - 8.25 = 3.75

The height would be the function (y = .125), or:
h = .125

Using the equation for the area of a rectangle: (A = h * w)

A = 3.75 * .125 = 0.469

Thus, P(8.25 < x < 12) = 0.469

3)
To find the probability GREATER than 10.50 pounds, we can subtract this value from the max value for the width:
W = 14 - 10.50 = 3.50

The height is the same as above:
h = .125

Solve for the area:
A = 3.50 * .125 = 0.4375

P(x > 10.5) = 0.4375

5 0
3 years ago
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