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2 years ago

14

The measure of angle 1 is (3x + 10)° In the measure of angle 4 is (4x - 15)°

1 answer:

netineya [11]2 years ago

8 0

∠1 = ∠4 ( vertically opposite angles )

Then :-

$3x + 10 = 4x - 15$

$10 = 4x - 15 - 3x$

$10 = 1x - 15$

$1x - 15 = 10$

$1x = 10 + 15$

$1x = 25$

Which means :-

∠1 =

= 3x + 10

= 3 × 25 + 10

= 75 + 10

= 85°

∠4 =

= 4x - 15

= 4 × 25 - 15

= 100 - 15

= 85°

Then :-

∠7 = ∠6

∠6 + ∠4 = 180° ( interior angles on the same side of transversal )

∠6 + 85 = 180°

∠6 = 180 - 85

∠6 = 95°

As ;

∠6 = ∠7

<h3>∠7 = 95°</h3>

<h2>Therefore , angle 7 = 95° .</h2>

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3 years ago

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15, -19, 23, -27.. how would you write that in sigma notation?

sertanlavr [38]

Answer: i think its zero

Step-by-step explanation:

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1 year ago

Please help me on this problem

Anna71 [15]

Answer:

The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

$a = -4, c = 5$
$a = 1, c = 6$
$a = 2, c = 3$
$a = 3, c = 3$

Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

$-4x^{2} -6x+5 = 0$ for $a = -4 , c=5$

$1x^{2} -6x+6 = 0$ for $a = 1 , c=6$

$2x^{2} -6x+3 = 0$ for $a = 2 , c=3$

$3x^{2} -6x+3 = 0$ for $a = 3 , c=3$

$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

$Discriminant = 116$ which is greater than 0 ∴ a real solution.

for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

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3 years ago

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Deffense [45]

Answer:

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Step-by-step explanation:

Use the reciprocal when dividing fractions. :)

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2 years ago