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Kay [80]
2 years ago
7

What is the slope and the y-intercept of the line on the graph below? On a coordinate plane, a line goes through points (0, 1) a

nd (4, 0). a slope = –4, y-intercept = 1 b slope = –4, y-intercept = 4 c slope = Negative one-fourth, y-intercept = 1 d slope = Negative one-fourth, y-intercept = 4
Mathematics
1 answer:
sashaice [31]2 years ago
6 0

Answer:

the slope of the coordinates will be 0-0/4-1 then you will get the coordinates of the graph

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what is the simplified form of the following expression? Assume x is greater than or equal to 0 and y is greater than or equal t
yarga [219]

the given expression is :

2(4√16x) - 2(4√2y) + 34√81x - 4(4√32y)

⇒ 8(√16x) - 8(√2y) + 34√81x - 16√32y  

⇒8×4√x - 8√2y + 34×9√x - 16√16×2y     [∵ √16 = 4 and √81 = 9]

⇒32√x - 8√2y + 306√x - 16×4√2y

⇒(32√x + 306√x) - 8√2y  - 16×4√2y      

⇒338√x -72√2y

4 0
3 years ago
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What is the inverse function of y = (x-4)^2 - 2​
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Answer:

Step-by-step explanation:

x = (y-4)^2 - 2

square root of x = y - 4 - 2

square root of x = y - 6

(square root of x) - 6 = y

4 0
3 years ago
All 3 questions 3 pictures for find the missing angles, please with reasoning why? 100 points
Katyanochek1 [597]

Answer:

Problem 1)

m\angle DEG = 38^\circ

Problem 2)

m\angle x =140^\circ \text{ and } m\angle y = 49^\circ

Problem 3)

m\angle EDG = 65^\circ

Step-by-step explanation:

Problem 1: ∠DEG

From the diagram, we know that ∠DFG intercepts Arc DG.

Inscribed angles have half the measure of its intercepted arc. Therefore:

\displaystyle \frac{1}{2}m\stackrel{\frown}{DG}=m\angle DFG

We know that m∠DFG = 38°. So:

\displaystyle \frac{1}{2}m\stackrel{\frown}{DG}=38\Rightarrow m\stackrel{\frown}{DG}=76^\circ

∠DEG also intercepts Arc DG. Hence:

\displaystyle m\angle DEG=\frac{1}{2}m\stackrel{\frown}{DG}

We know that Arc DG measures 76°. Hence:

\displaystyle m\angle DEG =\frac{1}{2}\left(76\right)=38^\circ

Alternate Explanation:

Since ∠DEG and ∠DFG intercept the same arc, ∠DEG ≅ DFG. So, m∠DEG = m∠DFG = 38°.

Problem 2: Circle with Centre O

(Let the bottom left corner be A, upper be B, and right be C.)

∠ACB intercepts Arc AC.

Inscribed angles have half the measure of its intercepted arc. Therefore:

\displaystyle \frac{1}{2}m\stackrel{\frown}{AC}=m \angle ACB

Since m∠ACB = 70°:

\displaystyle \frac{1}{2}m\stackrel{\frown}{AC}=70\Rightarrow m\stackrel{\frown}{AC}=140^\circ

∠<em>x</em> is a central angle and also intercepts Arc AC.

The measure of a central angle is equal to its intercepted arc. Thus:

\displaystyle m\angle x =m\stackrel{\frown}{AC}=140^\circ

The sum of the interior angles of a polygon is given by the formula:

(n-2)\cdot 180^\circ

Where <em>n</em> is the number of sides.

Since the inscribed figure is a four-sided polygon, its interior angles must total:

(4-2)\cdot 180^\circ =360^\circ

Therefore:

21+70+y+(360-x)=360

Substitute and solve for <em>y:</em>

91+y+(360-140)=360\Rightarrow y +311=360\Rightarrow m\angle y = 49^\circ

Problem 3: ∠EDG

∠EFG intercepts Arc EDG.

Inscribed angles have half the measure of its intercepted arc. Therefore:

\displaystyle \frac{1}{2}m\stackrel{\frown}{EDG}=m\angle EFG

Since m∠EFG = 115°:

\displaystyle \frac{1}{2}m\stackrel{\frown}{EDG} = 115\Rightarrow m\stackrel{\frown}{EDG} = 230^\circ

A full circle measures 360°. Hence:

m\stackrel{\frown}{EDG}+m\stackrel{\frown}{EFG}=360^\circ

Since we know that Arc EDG measures 230°:

230^\circ + m\stackrel{\frown}{EFG}=360

Solve for Arc EFG:

m\stackrel{\frown}{EFG}=130^\circ

∠EDG intercepts Arc EFG.

Inscribed angles have half the measure of its intercepted arc. Therefore:

\displaystyle m\angle EDG = \frac{1}{2}m\stackrel{\frown}{EFG}

Since we know that Arc EFG measures 130°:

\displaystyle m\angle EDG = \frac{1}{2}(130)=65^\circ

7 0
3 years ago
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13) Which fraction needs the fewest parts to mark 1 whole? <br>1/10,1/5,1/7,1/9​
ad-work [718]

Answer:

1/5

Step-by-step explanation:

it's denominator is 5, so it needs 5 parts to make a whole

3 0
3 years ago
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