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horsena [70]
3 years ago
5

What is the measure of angle TSU?

Mathematics
2 answers:
Elden [556K]3 years ago
6 0

Answer:

63°

Step-by-step explanation:

TSU = 180 - 117 = 63

barxatty [35]3 years ago
4 0

Answer:

plz give them brainliest

Step-by-step explanation:

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the temperature in Nico's town was -2.9 f at noon . the temperature increased at a steady rate of 1.3 f per hour until 6:00p.m.
kotykmax [81]
Temperature at noon: Tn=-2.9 f
Rate: r=+1.3 f/h
Total decreased: d=-3.7 f 
Temperature at midnight: Tm=?

Tm=Tn+r(6 p.m.-12 m)+d
Tm=-2.9 f+(1.3 f/h)(18 h-12 h)+(-3.7 f)
Tm=-2.9 f+(1.3 f/h)(6 h)-3.7 f
Tm=-2.9 f+7.8 f-3.7 f
Tm=1.2 f

Answer: T<span>he temperature at midnight was 1.2 f</span>
6 0
3 years ago
find two consecutive intergers who sum is equal to the difference of three times the larger and two times the smaller
vekshin1

Answer:

4 and 6

Step-by-step explanation:

Let integers be x and x+2 Then:-

x + x + 2  = 3(x + 2) - 2x

2x + 2 = 3x + 6 - 2x

2x + 2 = x + 6

x = 4

The numbers are 4 and 6.

3 0
2 years ago
100 POINTS NEED ANSWER ASAP
never [62]

Answer:

DOMAIN: {4,5,6,7}

RANGE: {4,4.5,5,5.5,6,6.5,7}

Step-by-step explanation:

here is proof

Mark Brainiest Please

4 0
2 years ago
Find the Area Of each circle.
Levart [38]

The area of the circle A and B is 86.54 inches and 124.62 mm

According to the statement

we have given that the radius of the circle A and b and we have to find the area of the given circle.

So, we know that the

The area enclosed by a circle of radius r is πr².

So, For area of the circle

For condition A :

diameter = 10.5 inches

then radius = 5.25

Area = πr²

Area = 3.14*27.56

Area = 86.54 inches

Now, For condition B :

radius = 6.3 mm

Area = πr²

Area = 3.14*39.69

Area = 124.62mm

So, The area of the circle A and B is 86.54 inches and 124.62 mm

Learn more about area of the circle here

brainly.com/question/15673093

#SPJ1

5 0
1 year ago
-1/225x^2+2/3x what is the vertex of this quadratic function?
Julli [10]
\bf \textit{vertex of a parabola}\\ \quad \\\\&#10;&#10;\begin{array}{lccclll}&#10;y=&-\frac{1}{225}x^2&+\frac{2}{3}x\\\\&#10;y=&-\frac{1}{225}x^2&+\frac{2}{3}x&+0\\&#10;&\uparrow &\uparrow &\uparrow \\&#10;&a&b&c&#10;\end{array}\qquad &#10;\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad  {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
6 0
3 years ago
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