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Papessa [141]
2 years ago
12

Help please

Mathematics
2 answers:
ratelena [41]2 years ago
6 0

Answer:

14 * 9(-9) - 8 = 1142               9 - 45(-3) + 7 = 151

Step-by-step explanation:

14 * 9(-9) - 8 = 1142               9 - 45(-3) + 7 = 151

(This one is true)                  (This one is false)

WINSTONCH [101]2 years ago
5 0

Answer:

57+20= 90 (false)

40+39= 79 (true)

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A box contains 18 green marbles and 8 white marbles. If the first marble chosen was a white marble, what is the probability of c
murzikaleks [220]

Answer:

7/25

Step-by-step explanation:

Original contents of the box:

18 green marbles

8 white marbles

Contents of the box after drawing 1 white marble:

18 green marbles

7 white marbles

Total: 25 marbles

p(white) = (number of white marbles)/(total number of marbles)

p(white) = 7/25

7 0
1 year ago
if a 42.9ft tall flagpole casts a 253.1ft long shadow then how long is the shadow that a 6.2ft tall woman casts?
vovangra [49]
Set up a proportion.
flagpole/shadow = woman/her shadow.

f/s1 = w/s2
f = 42.9
s1 = 253 
w = 6.2 
s2 = ??? = x

42.9/253 = 6.2 / x
42.9 x = 253*6.2
42.9 x = 1568.6
x = 1568.6 / 42.9
x = 36.56 The womans shadow is 36.56 feet long <<<< ===== answer.
3 0
3 years ago
Tim drew a triangle with one angle of measure 40 degrees and the other 70 degrees. Which triangle did Tim draw?
d1i1m1o1n [39]

Answer:

Isocoles (I can't spell)

Step-by-step explanation:

180-70=110. 110-40=70.

Two angles are 70

3 0
2 years ago
Multiples that are shared by two or more numbers are_______
Elanso [62]

Answer:

it is called Common Multiples

3 0
3 years ago
Read 2 more answers
In triangle ΔABC, ∠C is a right angle and CD is the height to
Zina [86]

Answer:

m\angle CDB=90\\m\angle CBD=90-\alpha\\m\angle BCD=\alpha\\\\m\angle CDA=90\\m\angle CAD=\alpha\\m\angle ACD=90-\alpha

Step-by-step explanation:

The triangles are drawn below.

CD is perpendicular to AB as CD is height to AB.

Therefore, angles m\angle CDB=m\angle CDA=90°

So, triangles ΔCBD and ΔCAD are right angled triangles.

Now, from the right angled triangle ΔABC,

m\angle A+m\angle B =90\\\alpha+m\angle B=90\\m\angle B=90-\alpha

From ΔCBD,

m\angle CBD is same as m\angle B.

So, m\angle CBD=90-\alpha

m\angle BCD+m\angle BDC =90\\m\angle BCD+90-\alpha=90\\m\angle BCD=\alpha

Now, from ΔCAD,

m\angle CAD is same as m\angle A

So, m\angle CAD=\alpha

m\angle CAD+m\angle ACD =90\\\alpha+m\angle ACD=90\\m\angle ACD=90-\alpha

Hence, the unknown angles of both the triangles are:

m\angle CDB=90\\m\angle CBD=90-\alpha\\m\angle BCD=\alpha\\\\m\angle CDA=90\\m\angle CAD=\alpha\\m\angle ACD=90-\alpha

5 0
2 years ago
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