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Law Incorporation [45]
3 years ago
5

Find the slope of the line​

Mathematics
2 answers:
PolarNik [594]3 years ago
8 0

Answer:

no

Step-by-step explanation:

ask a teacher so people might lie abt the answer

Mama L [17]3 years ago
3 0

Answer:

- 5/3

Step-by-step explanation:

This slope would be a positive slope since we have the dot on the negative side but it looks like it's in the ones but it's negative 5/3 in my opinion good luck

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In the triangle below, what is cos B?
Kruka [31]
Cos B = c/a

cosine is adjacent over hypotenuse.


there is an acronym for this: soh-cah-toa

soh: sine is opposite over hypotenuse
cah: cosine is adjacent over hypotenuse
toa: tangent is opposite over adjacent
7 0
4 years ago
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I’ll brainliest you please help me
Tanya [424]
X= 75
y= 105
this is because x is equal to the angle with 75, y us 105 because 180-75=105. 180 is found because the angle of a line is 180
3 0
3 years ago
60 POINTS I NEED HELPPPPPPPPPPP
GREYUIT [131]

<em>Refer to the attachment for solution and below for steps!!~</em>

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  • Write the number in exponential form with the base of 2/7
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3 0
2 years ago
A circle has its center at the center of a square with 26-inch sides. Using 3.14 for pi, find the area of the square not covered
AlladinOne [14]
For squrae
area=side²
perimiter=4(side)

for circle
area=pir²
circumference=2pir

side=diameter, diameter/2=radius=side/2


find area not covered
that is areasquare-areacircle
areasquare=26²=676 sqauare inches
areaciecle, diameter=side, 26/2=13, area=pi13²=169*3.14=530.66 square inches

areanotcovered=676-530.66=145.34 square inches not coverd


ratio of perimiter to circumferece
perimiter=4*26=104
circumference=2pir=2pi13=26pi=26*3.14=81.64
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7 0
3 years ago
Read 2 more answers
1. Determine the discriminant and nature of roots of each quadratic equation.
Fiesta28 [93]

Answer:

Please check the explanation.

Step-by-step explanation:

The general quadratic equation is

ax²+bx+c=0

The discriminant = D = b² - 4ac

When b² - 4ac = 0 there is one real root.

When b² - 4ac > 0 there are two real roots.

When b² - 4ac < 0 there are two complex roots.

1) x² -6x + 9=0

On comparing with given quadratic equation x² -6x + 9=0

a = 1, b=-6, c=9

D = b² - 4ac

   = (-6)² - 4(1)(9)

   = 36 - 36

   = 0

D = 0

Thus, there is one real root of quadratic equation x² -6x + 9=0.

2) x² -4x + 3=0

On comparing with given quadratic equation x² -4x + 3=0

a = 1, b=-4, c=3

D = b² - 4ac

   = (-4)² - 4(1)(3)

   = 16 - 12

   = 4

D > 0

Thus, there are two real roots of quadratic equation x² -4x + 3=0.

3)  x² -7x - 4=0

On comparing with given quadratic equation x² -7x - 4=0

a = 1, b=-7, c=-4

D = b² - 4ac

   = (-7)² - 4(1)(-4)

   = 49 + 16

   = 65

D > 0

Thus, there are two real roots of quadratic equation x² -7x - 4=0

4)  2x² +3x +5=0

On comparing with given quadratic equation 2x² +3x +5=0

a = 2, b=3, c=5

D = b² - 4ac

   = (3)² - 4(2)(5)

   = 9 - 40

   = -31

D < 0

Thus, there are two complex roots of quadratic equation 2x² +3x +5=0

8 0
3 years ago
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