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kolezko [41]
4 years ago
11

Need help with this question

Mathematics
1 answer:
tatuchka [14]4 years ago
4 0
Z(x)=6x^3+19x^2-52x+15
(6*-5)^3+19*-5^2-52*-5+15
Z(x)= -17,700
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4. Evaluate: (-7 - 41)(8 - 6) A) -32 + 10i C) -80 - 10i B) -80 - 10i D) -32 + 741
ycow [4]

Answer:

Step-by-step explanation:

(-7-41)(8-6)

-48*2

-96

This is what I got. I don't see anything in the equation that has "i" in it.

If it is supposed to be (-7-4i)(8-6) foil the problem.

-56+42-32i+24i  combine like terms

the answer is -14-8i

8 0
3 years ago
Order them least to greatest. And explain
AnnyKZ [126]

Answer:

13. 1/2, 5/8, 3/4

16. 7/12, 3/5, 2/3

Step-by-step explanation:

make them have the same denominator.

13) 1/2 x 4 = <u>4/8</u>, <u>5/8</u>, 3/4 x 2= <u>6/8</u>

16) Grab two fractions let's say 3/5 and 2/3

Multiply 3/5 denominator (5) with 2/3 numerator (2)

It's 10 and place it above the number you multiplied the numerator from so it's above 2/3.  Do it again but with 2/3 denominator (3) and 3/5 numerator (3). Multiply 3 and 3 you get 9. Place the 9 above 3/5. We know that 9 is less than 10 so 3/5 is less than 2/3. You can do this with any two fractions.

6 0
3 years ago
Evaluate g(x) = x2 + 5 given the inputs (-1,0,1,2}
mihalych1998 [28]

Answer:

g(-1) = 6

g(0) = 5

g(1) = 6

g(2) = 9

Step-by-step explanation:

Step 1: Plug in -1 for <em>x</em>

g(-1) = (-1)² + 5

g(-1) = 1 + 5

g(-1) = 6

Step 2: Plug in 0 for <em>x</em>

g(0) = 0² + 5

g(0) = 0 + 5

g(0) = 5

Step 3: Plug in 1 for <em>x</em>

g(1) = 1² + 5

g(1) = 1 + 5

g(1) = 6

Step 4: Plug in 2 for <em>x</em>

g(2) = 2² + 5

g(2) = 4 + 5

g(2) = 9

5 0
3 years ago
A worker drags a box of mass 50 kg across a horizontal floor. The worker attaches a rope to the box and pulls on the rope so tha
Arlecino [84]

Answer:

A) F=\displaystyle{\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)} }with a = \frac{2\Delta x}{t^2}

B) The magnitude of the pulling force is 24 % of the magnitude of the gravitational force acting on the box

Step-by-step explanation:

A) The forces acting on the box are the pulling force that the worker exerts on the rope, the friction, the normal force, and the gravitational force. See the attached diagram. Since the pulling force on the rope makes an angle with the horizontal, this will have components in the x and y-axis (see diagram).

In the y-axis, the box does not move, therefore:

\sum_{y} F = 0\\F\sin(\alpha) + N - F_g = 0\\N = F_g - F\sin(\alpha)\\N = mg - F\sin(\alpha) (1)

where \alpha is the given angle of the rope with the horizontal.

In the x-axis, the box moves with an acceleration a that can be calculated as a function of the given displacement and time interval as:

a = \frac{2\Delta x}{t^2} and the force equation is:

\sum_{x} F = ma\\F\cos(\alpha) -f =ma\\F\cos(\alpha) - \mu N = ma

now substituting N from equation (1):

F\cos(\alpha) - \mu N = ma\\F\cos(\alpha) - \mu (mg - F\sin(\alpha))=ma\\F\cos(\alpha) - \mu mg -\mu F\sin(\alpha)=ma\\F\cos(\alpha) -\mu F\sin(\alpha)=ma+ \mu mg\\F(\cos(\alpha) -\mu \sin(\alpha))=ma+ \mu mg\\\boxed{F=\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)}}

and putting the expression for the acceleration:

\boxed{F=\frac{m \frac{2\Delta x}{t^2}+ \mu mg}{\cos(\alpha) -\mu \sin(\alpha)}} (2)

which is the requested expression

B) Substituting the values in (2) and using g=9.8\ \rm{ms^{-2}}:

F=\frac{50 \cdot ( 0.762939+ 0.1\cdot \cdot 9.8)}{\cos(36.8^{\circ}) -0.1 \sin(36.8^{\circ})}\\F=\frac{87.147}{0.740829} \\F=11.634\ \rm{N}

The gravitational force acting on the box is:

F_g=mg=50\cdot9.8=490\ \rm{N}

F/F_g = (117.634/490)\cdot 100 = 24 \%

Thus, the magnitude of the pulling force is 24 % of the magnitude of the gravitational force acting on the box.

8 0
4 years ago
A sphere has a radius of 4x+4
andreyandreev [35.5K]
Ok then what is the question
4 0
4 years ago
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