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Goryan [66]
2 years ago
13

6-column table with 5 rows. The 1st column is labeled + x squared with entries +x, +x, +x, +x, +x. The 2nd column is labeled +x

with entries +, +, +, +, +. The 3rd column is labeled +x with entries +, +, +, +, +. The 4th column is labeled +x with entries +, +, +, +, +. The 5th column is labeled +x with entries +, +, +, +, +. The 6th column is labeled +x with entries +, +, +, +, +.
The algebra tiles represent the perfect square trinomial x2 + 10x + c.



What is the value of c?
Mathematics
1 answer:
denis-greek [22]2 years ago
5 0

Answer:

25

Step-by-step explanation:

5*5=25

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Graded Assignment Unit Test, Part 2: Basic Geometric Shapes Answer the questions below. When you are finished, submit this assig
AfilCa [17]

Answer:

In case (a) car makes 105° turn.

In case (b) car makes 75° turn.

In case (c) car makes 105° turn.

Step-by-step explanation:

Figure is redrawn To explain properly (in attachment)

Given : streets are parallel means \overline{AB} ║ \overline{CD},

             AB - 4th street , CD - 3rd street and XY - King Ave.

            ∠XLA = 75°

To find : (a) ∠XLB

              (b) ∠LMD (left onto 3rd streat means left of car)

              (c) ∠YMD (right means right side of car)

∠XLB + ∠XLA = 180° (Linear Pair = 2 adjacent angles are  

                                     supplementary)

∠XLB + 75° = 180°

∠XLB  = 180 - 75

∠XLB = 105°

∴ In case (a) car makes 105° turn.

∠LMD = ∠XLA = 75° (Corresponding angles of parallel lines are equal)

∠LMD = 75°

∴ In case (b) car makes 75° turn.

∠YMD + ∠LMD = 180° (Linear Pair = 2 adjacent angles are

                                       supplementary)

∠YMD + 75° = 180°

∠YMD = 180 - 75

∠YMD = 105°

∴In case (c) car makes 105° turn.

8 0
2 years ago
John's commute to work is 20kmhr while Sheri's commute is 500mmin. Who has the fastest commute to work in mihrif 1.61km=1mi? A S
Andreyy89

Answer:

A) Sheri has the faster commute by 6.2 miles/hr.

Step-by-step explanation:

Given

John's commute to work =20\ km/hr

Sheri's commute to work  =500\ m/min

1.61\ km = 1\ mile

John's commute to work in miles per hour = \frac{20\ km}{1 hr}\times \frac{1\ mile}{1.61\ km}= 12.42\ miles/hr

Sheri's commute to work in miles per hour =\frac{500\ m}{1\ min}\times \frac{1\ km}{1000\ m}\times \frac{1\ mile}{1.61\ km}\times \frac{60\ min}{1\ hr}= 18.63\ miles/ hr

We can see that Sheri has a faster commute.

Difference between the rates =18.63\ miles/ hr-12.42\ miles/hr=6.21\ miles/hr\approx 6.2\ miles/hr

∴ Sheri has the faster commute by 6.2 miles/hr.

3 0
2 years ago
A rectangular table is positioned in a 10-foot by 10-foot room as shown.
makkiz [27]

Answer:

40 and 20

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
100 POINTS!!!!ANSWER ASAP!!!
GarryVolchara [31]

2 mins per problem

10 mins and 5 questions left

hope it helps comment for any questions have an amazing day

6 0
2 years ago
Read 2 more answers
Suppose that the weight of seedless watermelons is normally distributed with mean 6.1 kg. and standard deviation 1.9 kg. Let X b
horrorfan [7]

Answer:

a. X\sim N(\mu = 6.1, \sigma = 1.9) b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349

Step-by-step explanation:

a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.

b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.

c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842

d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166

e. The z-score related to 6.4 kg is z_{1} = (6.4-6.1)/1.9 = 0.1579 and the z-score related to 7 kg is z_{2} = (7-6.1)/1.9 = 0.4737, we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194

f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349

7 0
2 years ago
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