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Arisa [49]
3 years ago
12

Product of 100 and sum of 25

Mathematics
1 answer:
ddd [48]3 years ago
6 0

Answer:

100*25

Step-by-step explanation:

sum of 25=25

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Lines e and f are parallel. The mAngle9 = 80° and mAngle5 = 55°. Parallel lines e and f are cut by transversal c and d. All angl
brilliants [131]

Answer:

(A)m\angle 2=125^\circ\\(C)m\angle 8=55^\circ\\(E)m\angle 14=100^\circ

Step-by-step explanation:

The diagram of the problem is drawn and attached.

Given that:

m\angle 9 =80^\circ\\m\angle 5 =55^\circ

m\angle 5+ m\angle 6=180^\circ\\55^\circ+ m\angle 6=180^\circ\\m\angle 6=180^\circ-55^\circ\\m\angle 6=125^\circ\\$Now:\\m\angle 6 = m\angle 2 $(Corresponging angles)\\Therefore m\angle 2=125^\circ\\

m\angle 5= m\angle 8=55^\circ $(Vertically Opposite Angles)

Also

m\angle 9+ m\angle 10=180^\circ\\80^\circ+ m\angle 10=180^\circ\\m\angle 10=180^\circ-80^\circ\\m\angle 10=100^\circ\\$Now:\\m\angle 10 = m\angle 14 $(Corresponging angles)\\Therefore m\angle 14=100^\circ\\

4 0
3 years ago
Read 2 more answers
Tori sells snow cones. His profit in dollars and high temperature in degrees Fahrenheit for two weeks are shown in the table bel
masya89 [10]
Sorry i cant answer if there is no table <3
6 0
3 years ago
I need it by tmr<br> ill give brainliest thx
Zina [86]

Answer:

21. B. Graph Z

23. A. neither function nor relation

24. D. relation only

Step-by-step explanation:

21. Graph Z is the only graph which does not pass the vertical line test

23. The graph does not pass the vertical line test, AND it is not merely given x and y points like a relation. It is graphed as a squiggly line with no given x and y points, therefore you could not make a relation from it either.

24. There is an x value of -4 twice, meaning the set does not pass the vertical line test, HOWEVER, it is a given collection of x and y values, making it a relation.

I would appreciate brainliest, have an amazing day

8 0
3 years ago
Triangle ABC has vertices A(-5, -2), B(7, -5), and C(3, 1). Find the coordinates of the intersection of the three altitudes
Darina [25.2K]

Answer:

Orthocentre (intersection of altitudes) is at (37/10, 19/5)

Step-by-step explanation:

Given three vertices of a triangle

A(-5, -2)

B(7, -5)

C(3, 1)

Solution A by geometry

Slope AB = (yb-ya) / (xb-xa) = (-5-(-2)) / (7-(-5)) = -3/12 = -1/4

Slope of line normal to AB, nab = -1/(-1/4) = 4

Altitude of AB = line through C normal to AB

(y-yc) = nab(x-xc)

y-1 = (4)(x-3)

y = 4x-11           .........................(1)

Slope BC = (yc-yb) / (xc-yb) = (1-(-5) / (3-7)= 6 / (-4) = -3/2

Slope of line normal to BC, nbc = -1 / (-3/2) = 2/3

Altitude of BC

(y-ya) = nbc(x-xa)

y-(-2) = (2/3)(x-(-5)

y = 2x/3 + 10/3 - 2

y = (2/3)(x+2)    ........................(2)

Orthocentre is at the intersection of (1) & (2)

Equate right-hand sides

4x-11 = (2/3)(x+2)

Cross multiply and simplify

12x-33 = 2x+4

10x = 37

x = 37/10  ...................(3)

substitute (3) in (2)

y = (2/3)(37/10+2)

y=(2/3)(57/10)

y = 19/5  ......................(4)

Therefore the orthocentre is at (37/10, 19/5)

Alternative Solution B using vectors

Let the position vectors of the vertices represented by

a = <-5, -2>

b = <7, -5>

c = <3, 1>

and the position vector of the orthocentre, to be found

d = <x,y>

the line perpendicular to BC through A

(a-d).(b-c) = 0                          "." is the dot product

expanding

<-5-x,-2-y>.<4,-6> = 0

simplifying

6y-4x-8 = 0 ...................(5)

Similarly, line perpendicular to CA through B

<b-d>.<c-a> = 0

<7-x,-5-y>.<8,3> = 0

Expand and simplify

-3y-8x+41 = 0 ..............(6)

Solve for x, (5) + 2(6)

-20x + 74 = 0

x = 37/10  .............(7)

Substitute (7) in (6)

-3y - 8(37/10) + 41 =0

3y = 114/10

y = 19/5  .............(8)

So orthocentre is at (37/10, 19/5)  as in part A.

8 0
3 years ago
Hi what is 4x3-1 x20=<br><br><br><br><br><br> Don’t mind this
krek1111 [17]

Answer:

4 x 3 - 1 x 20 = -8

H u m a n F i n g e r P i e.

3 0
3 years ago
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