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FromTheMoon [43]

2 years ago

6

Graph the function r(x) = x + 5

1 answer:

Gwar [14]2 years ago

7 0

Slope:

$1$

y-intercept:

(0,5)

HopeItHelps you

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6

Step-by-step explanation:

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It all depends on the fraction.

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a garden table and a bench cost $1,025 combined the garden table cost $75 more than the bench what is the cost of the bench

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2 years ago

Plot the points on a graph.A(0,0);B(0,2);C(2,2);D(2,0). Join AB, BC,CD and AD. What is the figure formed? Find the area of the f

tia_tia [17]

$\large \green{ \underline{ \blue{ \boxed{\bf{ \red{Given -}}}}}}$

Four points which lie at (0,0) ; (0,2) ; (2,2) ; (2,0)

$\large \green{ \underline{ \blue{ \boxed{\bf{ \red{To \: Find-}}}}}}$

What is the figure formed?What is the area of the figure?

$\large \green{ \underline{ \blue{ \boxed{\bf{ \red{Solution-}}}}}}$

The given points lie on graph as per the order given in the attachment.

Which shape?

As the points lie 2 units apart from their adjacent sides. It can be a square or rhombus.

The opposite pairs of sides are also parallel to each other. The adjacent sides of the figure formed are also perpendicular to each other.

<u>Hence! the figure formed is a square.</u>

What is the area?

$\large{\sf{ \longmapsto Area_{(square)} = s \times s}}$

$\large{\sf{ \longmapsto Area_{(square)} = {s}^{2} }}$

$\large{\sf{ \longmapsto Area_{(square)} = {(2)}^{2} }}$

$\large{\sf{ \longmapsto Area_{(square)} = 4 \: sq. \: units}}$

4 0

1 year ago

Which expression is equivalent to sqrt8x^7y^8? assume x>0

Ilia_Sergeevich [38]

Answer:

$\sqrt{8x^7y^8$ $= 2x^3y^4 * \sqrt{2x}$

Step-by-step explanation:

Given

$\sqrt{8x^7y^8$

Required

Evaluate

$\sqrt{8x^7y^8$

Separate each factor with multiplication sign

$\sqrt{8*x^7*y^8$

Express 8 as 4 * 2 and $x^7$ as $x^6 * x$

$\sqrt{4 * 2 *x^6 * x*y^8$

Split each factor

$\sqrt{4} * \sqrt{2} *\sqrt{x^6} * \sqrt{x}*\sqrt{y^8}$

Evaluate $\sqrt4$

$2 * \sqrt{2} *\sqrt{x^6} * \sqrt{x}*\sqrt{y^8}$

Apply law of indices:

$2 * \sqrt{2} * x^{6*\frac{1}{2}} * \sqrt{x} * y^{8*\frac{1}{2}}$

$2 * \sqrt{2} * x^3 * \sqrt{x} * y^4$

Reorder

$2* x^3 * y^4 * \sqrt{2} * \sqrt{x}$

$2x^3y^4 * \sqrt{2} * \sqrt{x}$

$2x^3y^4 * \sqrt{2*x}$

$2x^3y^4 * \sqrt{2x}$

$2x^3y^4 \sqrt {2x$

Hence:

$\sqrt{8x^7y^8$ $= 2x^3y^4 * \sqrt{2x}$

8 0

2 years ago

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