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3 years ago

Find the slope of a line through the points (-1,-9) and (4,-2).

Mathematics

2 answers:

sveticcg [70]3 years ago

5 0

The answer is 7/5

please see the attached picture for full solution

Hope it helps

Good luck on your assignment

Harrizon [31]3 years ago

3 0

Answer:

UP 7 OVER 5.( FORM POINT ( -1,-9))

Step-by-step explanation:

PLOT BOTH POINTS ON A GRAPH THAN JUST COUNT UP AND THEN OVER TILL YOU MEET THE OTHER POINT

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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°:

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

∠x 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 n 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 y:

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

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°:

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

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3 years ago

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HELP ASAP PLS i’ll really appreciate it!!

Airida [17]

Sorry this is sloppy

3 0

3 years ago