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astra-53 [7]
3 years ago
12

Help this is about shapes

Mathematics
1 answer:
denis-greek [22]3 years ago
8 0

Answer:

exactly one pair of parallel sides =E

two pairs of sices of equal length = B,D

no right angle = A,C

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Find the value of Y! please help
Maurinko [17]

Answer:

y = 5.6

Step-by-step explanation:

Use Sine

Sine = \frac{opposite}{hypothenuse}

Sine 45 = \frac{y}{18}

Sine of 18 is 0.309

0.309 = \frac{y}{18}

Multiply 18 on both sides

0.309 x 18 = \frac{y}{18} x 18

5.562 = y

Round to the nearest tenth:

5.562 = 5.6

3 0
3 years ago
Which statements accuratley describe the function f(x)=3(16)^3/4x
Varvara68 [4.7K]

Answer:

line

x-intercept | 0

f-intercept | 0

normal vector | (-3072/sqrt(9437185), 1/sqrt(9437185))≈(-1., 0.000325521)

slope | 3072

curvature | 0

Step-by-step explanation:

7 0
3 years ago
Super easy... #10-12. LOOK AT PICTURE! URGENT!
alisha [4.7K]
10
sum is add
cube is to the 3rd powr
19+x³


11.
product means multiply
3 times y is no more than (equal to or less than ) 21
3y≤21
no need to solve

12. 2 times difference is 10
difference is subtract

2(z-12)=10
7 0
3 years ago
If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by
Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by y = 70t-16t^2.

To find : The average velocity for the time period beginning when t = 2 and lasting.  a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :    

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}

(\text{Average velocity})_{0.1\ s}=696\ ft/s

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}

(\text{Average velocity})_{0.01\ s}=7536\ ft/s

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}  

(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}

(\text{Average velocity})_{0.001\ s}=75936\ ft/s

5 0
3 years ago
4(-9x + 1) =<br> Answer?
faltersainse [42]

Answer:

−36+4

Step-by-step explanation:

8 0
3 years ago
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