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Harlamova29_29 [7]
4 years ago
9

The sum of two consecutive integers is 99, what is the 2nd integer??

Mathematics
1 answer:
sertanlavr [38]4 years ago
7 0
The two consecutive numbers are 40 and 59.
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A system of equations is shown.<br> y = 3x - 2<br> y=x2
Yuki888 [10]

Answer:

x = 2 and x = 1

Step-by-step explanation:

I think

5 0
3 years ago
Divide 190 to the ratio 0.2:0.3:0.5
Lostsunrise [7]
0.2 + 0.3 + 0.5 = 1
190 x 0.2/1 = 38
190 x 0.3/1 = 57
190 x 0.5/1 = 95
   190
0.2: 0.3: 0.5
38: 57: 95
3 0
3 years ago
Please can you give me help with answers. Thanks
Margaret [11]
1) 110+30+a = 180
a = 180-140 = 40
Obtuse angled triangle

2) 50+50+C = 180
C=180-100 =80
Acute angled triangle

3) 45+45+b =180
b =180-90 = 90 
Right angled triangle

4) 45+60+C = 180
C = 180-105 =75 
Acute angled triangle

5) 94+47+b =180
b = 180-141=39

Obtuse angled triangle

6)81+53+a = 180
a =180-134 = 46

Acute angled triangle

7) 38+45+b =180
b = 180-83 =97
Obtuse angled triangle

8) 36+54+C =180
C =180-90 = 90

Right angled triangle.
5 0
3 years ago
A car dealership sold 480 vehicles last year. Of these vehicles, 15% were minivans.
natulia [17]
You can use my photo as reference but its going to be 15/100 of 480

7 0
3 years ago
Plz hurry!!!! thank you!!!!
Nikitich [7]

Answer:

Area of trapezium = 4.4132 R²

Step-by-step explanation:

Given, MNPK is a trapezoid

MN = PK and ∠NMK = 65°

OT = R.

⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).

Now, sum of interior angles in a quadrilateral of 4 sides = 360°.

⇒ x + x + 65° + 65° = 360°

⇒ x = 115°.

Here, NS is a tangent to the circle and ∠NSO = 90°

consider triangle NOS;

line joining O and N bisects the angle ∠MNP

⇒ ∠ONS = \frac{115}{2} = 57.5°

Now, tan(57.5°) = \frac{OS}{SN}

⇒ 1.5697 = \frac{R}{SN}

⇒ SN = 0.637 R

⇒ NP = 2×SN = 2× 0.637 R = 1.274 R

Now, draw a line parallel to ST from N to line MK

let the intersection point be Q.

⇒ NQ = 2R

Consider triangle NQM,

tan(∠NMQ) = \frac{NQ}{QM}

⇒ tan65° = \frac{NQ}{QM}

⇒ QM = \frac{2R}{2.1445}

QM = 0.9326 R .

⇒ MT = MQ + QT

          = 0.9326 R + 0.637 R  (as QT = SN)

⇒ MT = 1.5696 R

⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R

Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).

⇒ A = (\frac{NP + MK}{2}) × (ST)

       = (\frac{1.274 R + 3.1392 R}{2}) × 2 R

       = 4.4132 R²

⇒ Area of trapezium = 4.4132 R²

5 0
4 years ago
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