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Bas_tet [7]
3 years ago
10

DOES THIS SET OF ORDERED PAIRS REPRESENT A FUNCTION? {(2,4),(2,5), (2,6), (2,7) }

Mathematics
1 answer:
vlabodo [156]3 years ago
7 0
Answer:
No, this set of ordered pairs does not represent a function.

Explanation:
Functions have one y-value for a given x-value.

Ordered pairs are written (x, y).

In this set, when x=2, it has four y-values (4, 5, 6 and 7), so it cannot be a function.
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Name The postulate or theorem you can use to prove the triangles congruent
Zielflug [23.3K]
The answer is HL. HL is a theorem stating that if the hypotenuse and one leg are congruent to another pair of hypotenuse and leg then the triangles themselves are congruent. Hope this helps. 
8 0
3 years ago
Please help me someone
katovenus [111]

Answer: Surface area is 169π

Step-by-step explanation:

A=\pi \cdot \:d^2

A=\pi d^2

A=\pi 13^2

A=169\pi

6 0
3 years ago
There’s this one as well
Doss [256]

Answer:

V= area of cross-section x length

V = 0.5x(8x9) x 11

V = 396cm^3

Hope this helps!

6 0
3 years ago
How to solve this problem​
nevsk [136]

Let C be the center of the circle. The measure of arc VSU is 2+114x, so the measure of the minor arc VU is 360-(2+114x)=358-114x. The central angle VCU also has measure 358-114x.

Triangle CUV is isosceles, so the angles CVU and CUV are congruent. The interior angles of any triangle are supplementary (they add to 180 degrees) so

m\angle VCU+2m\angle CUV=180

\implies m\angle CUV=\dfrac{180-(358-114x)}2=57x-89

UT is tangent to the circle, so CU is perpendicular to UT. Angles CUV and VUT are complementary, so

(57x-89)+(31x+3)=90

\implies88x=176

\implies x=2

So finally,

m\widehat{VSU}=2+114\cdot2=230

degrees.

6 0
3 years ago
Find all solutions to
BARSIC [14]

Answer:

x= 0 , \frac{1}{14} , \frac{-1}{12}

Step-by-step explanation:

Given, equation is \sqrt[3]{15x-1} + \sqrt[3]{13x+1} = 4\sqrt[3]{x}. →→→ (1)

Now, by cubing the equation on both sides, we get

( \sqrt[3]{15x-1} + \sqrt[3]{13x+1} )³ = (4\sqrt[3]{x})³

⇒ (15x-1) + (13x+1) + 3×\sqrt[3]{15x-1}×\sqrt[3]{13x+1} (\sqrt[3]{15x-1} + \sqrt[3]{13x+1}) = 64 x.

⇒ 28x + 3×\sqrt[3]{15x-1}×\sqrt[3]{13x+1} (4\sqrt[3]{x}) = 64x.        

(since from (1),  \sqrt[3]{15x-1} + \sqrt[3]{13x+1} = 4\sqrt[3]{x})

⇒ 12× \sqrt[3]{15x-1}×\sqrt[3]{13x+1} (\sqrt[3]{x})= 36x.

⇒ 3x = \sqrt[3]{(15x-1)(13+1)(x)}.

Now, once again cubing on both sides, we get

(3x)³ = (\sqrt[3]{(15x-1)(13+1)(x)})³.

⇒ 27x³ = (15x-1)(13x+1)(x).

⇒ 27x³ = 195x³ + 2x² - x

⇒ 168x³ + 2x² - x = 0

⇒ x(168x² + 2x -1) = 0

⇒ by, solving the equation we get ,

x = 0 ; x = \frac{1}{14} ; x = \frac{-1}{12}

therefore, solution is x= 0 , \frac{1}{14} , \frac{-1}{12}

7 0
3 years ago
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