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3 years ago

What is the value of 10+/-3-6/

Mathematics

1 answer:

garri49 [273]3 years ago

7 0

Answer:

Step-by-step explanation:

-3-6= -8 but since its absolute value it'll be 8 so 8 plus 10 = 18

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A teacher wants to buy supplies for two of their students, Susie and John. This teacher wants to spend at least $5 on each stude

tino4ka555 [31]

X and y represent the two students.

Teacher wants to spend at least $5 in each of x and y. That means x or y could be either equal to 5 or higher than 5.
x ≥ 5
y ≥ 5

Teacher only spends under $30. That means the sum of x and y couldn't be equal to higher than 30. It should be lower than 30.
x + y < 30

The correct answer is option D

6 0

3 years ago

Need help with all of this please

Harrizon [31]

Answer:

Step-by-step explanation:

In a right angled triangle, we have perpendicular, hypotenuse and base.

The hypotenuse is the longest side and opposite to the right angle. the side having 90 degree angle is perpendicular.

Applying formulas we can find the values:

the formulas are : cos (Ф) = Base / hypotenuse

sin (Ф) = Perpendicular / hypotenuse

tan (Ф) = Perpendicular / Base

11. cos z

cos z = Base / hypotenuse

cos z = 12/15

12. cos C

cos C = base / hypotenuse

cos C = 38/45

13. tan C

tan C = Perpendicular/ Base

tan C = 40/30

14. tan A

tan A = Perpendicular/ Base

tan A = 21/20

15. tan C

tan C = Perpendicular/ Base

tan C = 12/35

16. tan X

tan X = Perpendicular/ Base

tan X = 30/40

17. sin Z

Sin Z = Perpendicular / Hypotenuse

sin Z = 35/37

18. sin z = Perpendicular / Hypotenuse

sin z = 30/50

8 0

3 years ago

Find the measurements (the length L and the width W) of an inscribed rectangle under the line with the 1st quadrant of the x &am

Leni [432]

The question is incomplete. Here is the complete question.

Find the measurements (the lenght L and the width W) of an inscribed rectangle under the line y = - $\frac{3}{4}$ x + 3 with the 1st quadrant of the x & y coordinate system such that the area is maximum. Also, find that maximum area. To get full credit, you must draw the picture of the problem and label the length and the width in terms of x and y.

Answer: L = 1; W = 9/4; A = 2.25;

Step-by-step explanation: The rectangle is under a straight line. Area of a rectangle is given by A = L*W. To determine the maximum area:

A = x.y

A = x(- $\frac{3}{4}.x + 3$ )

A = - $\frac{3}{4}.x^{2} + 3x$

To maximize, we have to differentiate the equation:

$\frac{dA}{dx}$ = $\frac{d}{dx}$ (- $\frac{3}{4}.x^{2} + 3x$ )

$\frac{dA}{dx}$ = -3x + 3

The critical point is:

$\frac{dA}{dx}$ = 0

-3x + 3 = 0

x = 1

Substituing:

y = - $\frac{3}{4}$ x + 3

y = - $\frac{3}{4}$ .1 + 3

y = 9/4

So, the measurements are x = L = 1 and y = W = 9/4

The maximum area is:

A = 1 . 9/4

A = 9/4

A = 2.25

6 0

3 years ago

My homework lesson 2 order of operations

djyliett [7]

Not sure what your question is but if you are learning the order of operations you can use the phrase "Please Excuse My Dear Aunt Sally"....

Parentheses
Exponents
Multiplication from left to right
Division from left to right
Addition from left to right
Subtraction from left to right

Hope this help :)

3 0

2 years ago

The equation 2x2 − 12x + 1 = 0 is being rewritten in vertex form. Fill in the missing step. Given 2x2 − 12x + 1 = 0 Step 1 2(x2

Bezzdna [24]

Answer:

Part 1) $2(x-3)^{2}-17=0$ (the missing steps in the explanation)

Part 3) (8, 4); The vertex represents the maximum profit

Part 4) x = 3.58, 0.42

Part 5) x = 6, x = 44; The zeros represent the number of monthly memberships where no profit is made

Part 6) 2(x − 7)2 + 118; x = $7

Part 7) The maximum height of the puck is 4 feet. −(x − 4)^2 + 6

Part 8) (x + 3)^2 − 4

Part 9) 2(x − 1)^2 = 4

Part 10) 8(x − 4)^2 + 592

Step-by-step explanation:

Part 1) we have

$2x^{2} -12x+1=0$

Convert to vertex form

step 1

Factor the leading coefficient and complete the square

$2(x^{2} -6x)+1=0$

$2(x^{2} -6x+9)+1-18=0$

step 2

$2(x^{2} -6x+9)+1-18=0$

$2(x^{2} -6x+9)-17=0$

step 3

Rewrite as perfect squares

$2(x-3)^{2}-17=0$

Part 3) we have

$f(x)=-x^{2}+16x-60$

we know that

This is the equation of a vertical parabola open downward

The vertex is a maximum

Convert to vertex form

$f(x)+60=-x^{2}+16x$

Factor the leading coefficient

$f(x)+60=-(x^{2}-16x)$

Complete the squares

$f(x)+60-64=-(x^{2}-16x+64)$

$f(x)-4=-(x^{2}-16x+64)$

Rewrite as perfect squares

$f(x)-4=-(x-8)^{2}$

$f(x)=-(x-8)^{2}+4$

The vertex is the point (8,4)

The vertex represent the maximum profit

Part 4) Solve for x

we have

$-2(x-2)^{2}+5=0$

$-2(x-2)^{2}=-5$

$(x-2)^{2}=2.5$

square root both sides

$(x-2)=(+/-)1.58$

$x=2(+/-)1.58$

$x=2(+)1.58=3.58$

$x=2(-)1.58=0.42$

Part 5) we have

$f(x)=-x^{2}+50x-264$

we know that

The zeros or x-intercepts are the value of x when the value of the function is equal to zero

In this context the zeros represent the number of monthly memberships where no profit is made

To find the zeros equate the function to zero

$-x^{2}+50x-264=0$

$-x^{2}+50x=264$

Factor -1 of the leading coefficient

$-(x^{2}-50x)=264$

Complete the squares

$-(x^{2}-50x+625)=264-625$

$-(x^{2}-50x+625)=-361$

$(x^{2}-50x+625)=361$

Rewrite as perfect squares

$(x-25)^{2}=361$

square root both sides

$(x-25)=(+/-)19$

$x=25(+/-)19$

$x=25(+)19=44$

$x=25(-)19=6$

Part 6) we have

$-2x^{2}+28x+20$

This is a vertical parabola open downward

The vertex is a maximum

Convert the equation into vertex form

Factor the leading coefficient

$-2(x^{2}-14x)+20$

Complete the square

$-2(x^{2}-14x+49)+20+98$

$-2(x^{2}-14x+49)+118$

Rewrite as perfect square

$-2(x-7)^{2}+118$

The vertex is the point (7,118)

therefore

The video game price that produces the highest weekly profit is x=$7

Part 7) we have

$f(x)=-x^{2}+8x-10$

Convert to vertex form

$f(x)+10=-x^{2}+8x$

Factor -1 the leading coefficient

$f(x)+10=-(x^{2}-8x)$

Complete the square

$f(x)+10-16=-(x^{2}-8x+16)$

$f(x)-6=-(x^{2}-8x+16)$

Rewrite as perfect square

$f(x)-6=-(x-4)^{2}$

$f(x)=-(x-4)^{2}+6$

The vertex is the point (4,6)

therefore

The maximum height of the puck is 4 feet.

Part 8) we have

$x^{2}+6x+5$

Convert to vertex form

Group terms

$(x^{2}+6x)+5$

Complete the square

$(x^{2}+6x+9)+5-9$

$(x^{2}+6x+9)-4$

Rewrite as perfect squares

$(x+3)^{2}-4$

Part 9) we have

$2x^{2}-4x-2=0$

This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 2 the leading coefficient

$2(x^{2}-2x)-2=0$

Complete the square

$2(x^{2}-2x+1)-2-2=0$

$2(x^{2}-2x+1)-4=0$

Rewrite as perfect squares

$2(x-1)^{2}-4=0$

$2(x-1)^{2}=4$

The vertex is the point (1,-4)

Part 10) we have

$8x^{2}-64x+720$

This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 8 the leading coefficient

$8(x^{2}-8x)+720$

Complete the square

$8(x^{2}-8x+16)+720-128$

$8(x^{2}-8x+16)+592$

Rewrite as perfect squares

$8(x-4)^{2}+592$

the vertex is the point (4,592)

The population has a minimum at x=4 years ( that is after 4 years since 1998 )

6 0

2 years ago