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Ratling [72]
3 years ago
11

At the space center, Karen bought a model of a shuttle. She started working on the model the next day at 11:13 A.M. She worked u

ntil leaving for lunch at 11:54 A.M. After lunch, she worked on the model again from 1:29 P.M. Until 1:46 P.M. How long did Karen work on the model? Karen worked on the model for minutes.
Mathematics
1 answer:
goblinko [34]3 years ago
3 0

Answer:

58 minutes.

Step-by-step explanation:

From the question, Karen worked on the model in two windows;

1) 11.13 AM - 11.54 AM = 41 minutes

2) 1.29 PM - 1.46 PM = 17 minutes.

Hence, the total amount of time Karen spent working on the Shuttle model is:

41minutes + 17minutes = 58 minutes.

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What is the value of x in the equation 4(2x + 10) = 0? (1 point)
miskamm [114]

Answer:

x = -5

Explanation:

4(2x + 10) = 0

[ Simplify both sides of the equation ]

4(2x + 10) = 0

(4)(2x) + (4)(10) = 0 [Distribute]

8x + 40 = 0

[ Subtract 40 from both sides ]

8x + 40 − 40 = 0 − 40

8x = −40

[ Divide both sides by 8 ]

8x  / 8 = −40 / 8

x = -5

4 0
3 years ago
Find <br><br>dr/ds if r = s³/2 +1<br>​
DIA [1.3K]

Answer:

hniuhbyutrdewsdrtfvbgyi

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Which answer choice do I choose?
Ratling [72]
The second answer is correct because in order for the ordered pairs to represent a function, there can only be 1 y value for every x value. The second option has 2 different y values for x = 3 so it is not a function.
The answer is the second option.
3 0
3 years ago
Calculate the area of the triangle with the following vertices (3, -7), (6, 4), (-2, -3)
Monica [59]

Answer:

\boxed{\mathsf{A} \triangle = \red{\dfrac{67}{2}u.a}}

Step-by-step explanation:

Let's follow up with the solution. Considering a triangle with the vertices \mathsf{A(x_A, y_A)}, \mathsf{B(x_B, y_B)} and \mathsf{C(x_C, y_C)}, have a look at the representation in the cartesian plan.

From this representation we can say that the area (A) of a triangle through the knowledge of <u>analytical geometry</u> is given by the determinant of the vertices divided by two, mathematically,

\mathsf{A} \triangle =  \dfrac{\left| \begin{array}{ccc}  \mathsf{x_A} & \mathsf{y_A }& 1 \\  \mathsf{x_B} &  \mathsf{ y_B} & 1 \\ \mathsf{ x_C} &  \mathsf{ y_C} & 1 \end{array} \right|}{2}

So, applying this knowledge we're going to have,

\mathsf{A} \triangle =  \dfrac{\left| \begin{array}{ccc}  3 & -7 & 1 \\ 6 &  4 & 1 \\ -2 &  -3 & 1 \end{array} \right|}{2}

\mathsf{A} \triangle =  \dfrac{1}{2}\left[  \left.\begin{array}{ccc}   3 & -7 & 1 \\ 6 &  4 & 1 \\ -2&  -3 & 1 \end{array}  \right| \begin{array}{cc} 3 & -7 \\ 6 & 4 \\ -2 & -3 \end{array} \right]

\mathsf{A} \triangle = \dfrac{12 + 14 - 18 - (-8 - 9 - 42)}{2}

\red{\mathsf{A} \triangle = \dfrac{67}{2} = 33,5u.a}

Hope you enjoy it, see ya!)

\green{\mathsf{FROM}}: Mozambique, Maputo – Matola City – T-3

DavidJunior17

3 0
3 years ago
I already tried C=pi times diameter but it didn't work so can someone help me please.
slamgirl [31]
C=2pi(diameter)
C=2pi(14)
C=28pi
8 0
3 years ago
Read 2 more answers
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