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Molodets [167]
4 years ago
5

Which is true? 11.949>12.00 14.887<14.097

Mathematics
2 answers:
SIZIF [17.4K]4 years ago
7 0
Niether
hope this helps!!!
zmey [24]4 years ago
6 0

Answer:

\large \boxed{\sf{Neither \ are \ right}}

Step-by-step explanation:

11.949 is not greater than 12.00.

11.949 < 12.00

14.887 is not less than 14.097.

14.887 > 14.097

You might be interested in
Philipe
UkoKoshka [18]
<h3>Philipe has to make a weekly sales of $ 3000 so that he can earn $ 360</h3>

<em><u>Solution:</u></em>

Given that,

Philipe  works for a computer store that pays a 12% commission and no salary

We have to find his weekly sales have to be for him to earn $360

From given, we can say,

12 % of weekly sales will earn him $ 360

12 % of weekly sales = 360

Let "x" be the weekly sales

12 % of "x" = 360

Solve the above expression

\frac{12}{100} \times x = 360\\\\0.12x = 360\\\\x = \frac{360}{0.12}\\\\x = 3000

Thus Philipe has to make a weekly sales of $ 3000 so that he can earn $ 360

6 0
4 years ago
Do you go for a run on Monday. You run 1. 23 km. On Tuesday you run the same distance. how far did you run all together?
Troyanec [42]
Add it together and find the answer
7 0
3 years ago
Read 2 more answers
Find an equation of the plane with the given characteristics. The plane passes through the points (4, 3, 1) and (4, 1, -7) and i
Minchanka [31]

Answer:

3x - 4y + z = 1

Step-by-step explanation:

Given

Point\ 1 = (4,3,1)

Point\ 2 = (4,1,-7)

Perpendicular to 8x + 7y + 4z = 18

Required

Determine the plane equation

The general equation of a plane is:

a(x-x_1) + b(y - y_1) + c(z-z_1) = 0

For n =

(x_1,y_1,z_1) = (4,3,1)

(x_2,y_2,z_2) = (4,1,-7)

First, we need to determine parallel vector V_1

V_1 =

V_1 =

V_1 =

V_1 is parallel to the required plane

From the question, the required plane is perpendicular to 8x + 7y + 4z = 18

Next, we determine vector V_2

V_2 =

This implies that the required plane is parallel to V_2

Hence: V_1 and V_2 are parallel.

So, we can calculate the cross product V_1 * V_2

V_1 =

V_2 =

n = V_1 * V_2

V_1 * V_2 =\left[\begin{array}{ccc}i&j&k\\0&2&8\\8&7&4\end{array}\right]

The product is always of the form + - +

So:

V_1 * V_2 = i\left[\begin{array}{cc}2&8\\7&4\end{array}\right]  -j\left[\begin{array}{cc}0&8\\8&4\end{array}\right] +k\left[\begin{array}{cc}0&2\\8&7\end{array}\right]

Calculate the product

V_1 * V_2 = i(2*4- 8*7) - j(0*4- 8*8) + k(0*7 - 2 * 8)

V_1 * V_2 = i(8- 56) - j(0- 64) + k(0 - 16)

V_1 * V_2 = i(-48) - j(- 64) + k(- 16)

V_1 * V_2 = -48i +64j - 16k

So, the resulting vector, n is:

n =

Recall that:

n =

By comparison:

a = -48   b = 64   c = -16

Substitute these values in a(x-x_1) + b(y - y_1) + c(z-z_1) = 0

-48(x-x_1) + 64(y - y_1) -16(z-z_1) =0

Recall that:(x_1,y_1,z_1) = (4,3,1)

So, we have:

-48(x-4) + 64(y - 3) -16(z-1) =0

-48x + 192 + 64y -192 - 16z +16 = 0

Collect Like Terms

-48x + 64y - 16z = 0 - 192 + 192 - 16

-48x + 64y - 16z = -16

Divide through by -16

3x - 4y + z = 1

<em>Hence, the equation of the plane is</em>3x - 4y + z = 1<em></em>

4 0
3 years ago
What is a solution to the system of equations that includes quadratic function f(x) and linear function g(x)? f(x) = 2x2 + x + 4
olga_2 [115]

The solution to the system of equations that includes quadratic function f(x) and linear function g(x) is

(\sqrt{2},\sqrt{-2}  ) and ((7+\sqrt{2)} ,(7+\sqrt{-2}))

We have given that,

f(x) = 2x^2 + x + 4

x                          g(x)      

-2                               1      

  -1                              3        

  0                              5      

  1                              7

  2                              9

<h3>What is a polynomial function?</h3>

A polynomial function is a relation where a dependent variable is equal to a  polynomial expression.

A polynomial expression is an expression including numbers and variables, where variables are raised to non-negative powers.

The general form of a polynomial expression is:

a₀ + a₁x + a₂x² + a₃x³ + ... + anxⁿ.

The highest power to a variable is the degree of the polynomial expression. When degree = 2, the function is a quadratic function.

When degree = 1, the function is a linear function.

<h3>How do we solve the given question?</h3>

The quadratic function is given to us:f(x) = 2x^2 + x + 4.

We need to determine the linear equation g(x).

Since it's a linear equation we use the two-point method to determine the equation.

<h3>What is the two-point?</h3>

y-y₁ = ((y₂-y₁)/(x₂-x₁))*(x-x₁)

We take the points g(-2) =1, g(-1) = 3

g(x) - g(1) = ((g(-2)-g(-1))/(-2+1))*(x-1)or,

g(x) - 1 = ((-2-(-1))/(-2+1))*(x-1)or, g(x) - 1 = -1(-x+1)or,

g(x) = x - 1 + 1 = x

∴ g(x) = x , is the linear function g(x)

We are asked to find the solution to the system of equations f(x) and g(x).To find the solution we need to check what is the common solution to both f(x) and g(x).

For that, we equate f(x) and g(x).2x^2 + x + 4 = x or, 2x² - x +x- 4 = 0or, 2x^2  - 4 = 02(x^2-2)=0x^2-2=0x^2-\sqrt{2}=0(x-\sqrt{2}) (x+\sqrt{2})=0(x-\sqrt{2})=0 \\x=\sqrt{2}  or x=-\sqrt{2}g(-1) = 3(from the table)g(\sqrt{2})=\sqrt{2}  \ and \ g(\sqrt{-2}) =-\sqrt{2}f(\sqrt{2}) = 2(\sqrt{2} )^2 + (\sqrt{2} ) + 3\\=2(2)+\sqrt{2} +3\\=7+\sqrt{2}g(-\sqrt{2} )= 2(\sqrt{-2} )^2 + (\sqrt{-2} ) + 3\\=2(2)+\sqrt{-2} +3\\=7+\sqrt{-2}

The solution to the system of equations that includes quadratic function f(x) and linear function g(x) is (\sqrt{2},\sqrt{-2}  ) and ((7+\sqrt{2)} ,(7+\sqrt{-2}))

Learn more about linear and  quadratic equations at

brainly.com/question/14075672

#SPJ1

5 0
2 years ago
If f(x) is differentiable for the closed interval [-1, 4] such that f(-1) = -3 and f(4) = 12, then there exists a value c, -1&lt
Alex_Xolod [135]

Answer: Choice A,   f ' (c) = 3

============================================

Work Shown:

f(-1) = -3 means the point (-1,-3) is on the f(x) curve.

f(4) = 12 means (4,12) is on the f(x) curve as well.

Compute the slope of the line through those two points.

Slope formula

m = (y2 - y1)/(x2 - x1)

m = (12 - (-3))/(4 - (-1))

m = (12+3)/(4+1)

m = 15/5

m = 3

The slope of the secant line through (-1,-3) and (4,12) is m = 3

Through the mean value theorem (MVT), there exists at least one value c such that f ' (c) = 3, where -1 < c < 4, and f(x) is a continuous and differentiable function on this interval in question.

Visually, there exists at least one tangent line that has the same slope of the secant line mentioned. Lines with equal slopes, and different y intercepts, are parallel.

5 0
4 years ago
Read 2 more answers
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