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scZoUnD [109]
3 years ago
11

Pls help. i will give brainliest to the most helpful, detailed answer

Mathematics
2 answers:
Rasek [7]3 years ago
5 0
Number 3 using table is 0.59, 3.41
Graphed on calculator and changed the table set to increments of .1 all the way to .000001 to estimate where the zeros where.

Number 1,2,4,5 are displayed in attachment.

Nikolay [14]3 years ago
5 0

Answer:

the answers are

Step-by-step explanation:

1)a.-3,-8

2)c.-6, -5/2

3)A. 0.59, 3.41

4)D. 5,1

5)D. 3.74 seconds

quadratic equation quiz check unit 6 lesson 5

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947/7 it is hard because it is confused
Lerok [7]

Answer:

135.2857142857143

Step-by-step explanation:

its right?

4 0
3 years ago
Read 2 more answers
Pls help me with this question will give BRAINLIST to best answer and extra points
PolarNik [594]

7-1=1.64575

2/3 17 = 2(17)/3

3n/2= 4.71238

4 0
3 years ago
Consider randomly selecting a student at a large university. Let A be the event that the selected student has a Visa card, let B
Tems11 [23]

Answer:

0.75

Step-by-step explanation:

Given,

P(A) = 0.6, P(B) = 0.4, P(C) = 0.2,

P(A ∩ B) = 0.3, P(A ∩ C) = 0.12, P(B ∩ C) = 0.1 and P(A ∩ B ∩ C) = 0.07,

Where,

A = event that the selected student has a Visa card,  

B = event that the selected student has a MasterCard,  

C = event that the selected student has an American Express card,

We know that,

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)

= 0.6 + 0.4 + 0.2 - 0.3 - 0.12 - 0.1 + 0.07

= 0.75

Hence, the probability that the selected student has at least one of the three types of cards is 0.75.

5 0
3 years ago
5/6 + 3/8; round 5/6 to the closest benchmark.
blagie [28]
Make the denominators the same and then multiply the numerators and then add the fractions together . You should get 29/24
4 0
3 years ago
PLS HELP I WILL MARK BRAINLIEST
Daniel [21]

Answer:

Step-by-step explanation:

Here we are going to use the rule which says that

i) equal arc segments subtends equal angles at the circle

ii) The Angle subtended by any arc segment at center is double to that of the angle subtended by the same arc at its circumference.

For more details please refer to the image attached to this problem.

Let us say that the angle subtended by arc mAB at center O = 6∅

Hence , ∠AOB=6∅

Hence ∠ADB = 3∅ ( Rule ii as discussed above )

Also as length of arc mCD = x , the angle subtended by it on the center will be in the same ratio as it was subtended by arc with length 6x

Hence

∠COD=∅

Hence

∠CAD=∅/2

Hence in ΔATD

∠ATD + ∠ADT +∠DAT = 180°

∠ATD + 3∅+∅/2= 180°

∠ATD = 180° - (3∅+∅/2)    ----------(A)

Also

∠ATD + ∠ATB = 180°

From (A)

180° - (3∅+∅/2) +∠ATB =  180°

∠ATB = (3∅+∅/2)

∠ATB = (6∅+∅)/2

∠ATB = (7∅)/2

However , in order to find the exact value of∠ATB  we need to evaluate  ∅, and to find it , we must have some value of x .

3 0
3 years ago
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