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2 years ago

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Which expression has a conflict of 10 and a constant of5

1 answer:

Natalka [10]2 years ago

7 0

The expression that has a coefficient of 10 and constant of 5 is 10x + 5.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

A coefficient is a number or any symbol representing a constant value that is multiplied by a variable. A constant is a value or number that never changes in expression.

The expression that has a coefficient of 10 and constant of 5 is 10x + 5.

Find out more on equation at: brainly.com/question/2972832

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I’ve been stuck on this for 3 days today is the last day please help me guys

OleMash [197]

y = x^2 - 2x - 3

Factored Form: y = (x - 3)(x + 1)

X-Intercepts: (3, 0) and (-1, 0)

Axis of Symmetry: x = 1

Vertex: (1, -4)

Domain: All Real Numbers (or negative infinity to infinity)

Range: y > -4

Hope this helps!

7 0

2 years ago

What is the discriminante of 9x^2+10x

Leni [432]

The discriminant of y=ax^2+bx+c is b^2-4ac

given
9x^2+10x+0
a=9
b=10
c=0

discriminant=10^2-4(9)(0)=100-0=100

the discriminant is 100

6 0

3 years ago

Free_Kalibri [48]

Answer:

T(2, -7) = (4, -10)

Step-by-step explanation:

T(x, y) = (x + 2, y - 3)

T(2, -7) = (2 + 2, -7 - 3)

T(2, -7) = (4, -10)

3 0

3 years ago

A surf instructor has an initial fee of $12 and charges $8 per hour for lessons,

professor190 [17]

Answer:

I am stuck on that question tooo

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro

Ugo [173]

Answer:

$Side\ B = 6.0$

$\alpha = 56.3$

$\theta = 93.7$

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with $\alpha, \beta, \theta$

[See Attachment for Triangle]

$A = 10cm$

$C = 12cm$

$\beta = 30$

What the question is to calculate the third length (Side B) and the other 2 angles ( $\alpha\ and\ \theta$ )

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

$B^2 = A^2 + C^2 - 2ABCos\beta$

Substitute: $A = 10$ , $C =12$ ; $\beta = 30$

$B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30$

$B^2 = 100 + 144 - 240*0.86602540378$

$B^2 = 100 + 144 - 207.846096907$

$B^2 = 36.153903093$

Take Square root of both sides

$\sqrt{B^2} = \sqrt{36.153903093}$

$B = \sqrt{36.153903093}$

$B = 6.0128115797$

$B = 6.0$ <em>(Approximated)</em>

Calculating Angle $\alpha$

$A^2 = B^2 + C^2 - 2BCCos\alpha$

Substitute: $A = 10$ , $C =12$ ; $B = 6$

$10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 180 - 144 *Cos\alpha$

Subtract 180 from both sides

$100 - 180 = 180 - 180 - 144 *Cos\alpha$

$-80 = - 144 *Cos\alpha$

Divide both sides by -144

$\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}$

$\frac{-80}{-144} = Cos\alpha$

$0.5555556 = Cos\alpha$

Take arccos of both sides

$Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)$

$Cos^{-1}(0.5555556) = \alpha$

$56.25098078 = \alpha$

$\alpha = 56.3$ <em>(Approximated)</em>

Calculating $\theta$

Sum of angles in a triangle = 180

Hence;

$\alpha + \beta + \theta = 180$

$30 + 56.3 + \theta = 180$

$86.3 + \theta = 180$

Make $\theta$ the subject of formula

$\theta = 180 - 86.3$

$\theta = 93.7$

5 0

3 years ago