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3 years ago

HELP PLZZ TIMED TEST 5 MIN LEFT!

Mathematics

2 answers:

Alina [70]3 years ago

7 0

Step-by-step explanation:

answer: 135 degrees for the missing side

sdas [7]3 years ago

6 0

The fourth one is the answer

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Select the correct answer, What is the least common multiple of the numbers 5, 25, and 157 A 25 B, 50 C. 75 D 100 E, 125

yKpoI14uk [10]

$\begin{gathered} \text{Multiples of 5:} \\ 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80 \\ \text{Multiples of }25\colon \\ 25,50,75,100,125,150,175,200,225,250 \\ \text{Multiples of 15:} \\ 15,30,45,60,75,90,105,120,135,150 \\ \text{LCM}=75 \end{gathered}$

6 0

1 year ago

Use technology or a z-score table to answer the question. Scores on a standardized military exam are normally distributed with a

iVinArrow [24]

Approximately 3365 students will score less than 66.

The z-score is calculated using the formula
z=(X-μ)/σ, where μ is the mean and σ is the standard deviation.

For this problem, we have
z=(66-57)/9 = 9/9 = 1.00

Using a z-table (http://www.z-table.com) we see that the area to the left of, or probability less than, this is 0.8413.

To find the number of students out of 4000 that will score in this range, we multiply this probability by 4000:

0.8413(4000) = 3365.2 ≈ 3365

3 0

3 years ago

What is 132.6 in expanded form

Alika [10]

Answer:

Step-by-step explanation:

Expanded Notation Form:

100

0.6

Expanded Factors Form:

1 ×

100

3 ×

2 ×

6 ×

0.1

Expanded Exponential Form:

1 × 102

3 × 101

2 × 100

6 × 10-1

6 0

3 years ago

Read 2 more answers

how can you please help me with that Write a linear equation for the line going through the points (-5, -13) and (-2, 5) Write y

klasskru [66]

Answer:

y=6x+17

Step-by-step explanation:

y=6x+17

3 0

3 years ago

{ HARDER }

cestrela7 [59]

(i):
$x = Vtcos\theta$
$Vcos\theta = \frac{x}{t}$

$y = -\frac{1}{2}gt^{2} + Vtsin\theta$
$Vsin\theta = \frac{y + \frac{1}{2}gt^{2}}{t}$

$V^{2}cos^{2}\theta + V^{2}sin^{2}\theta = \frac{x^{2}}{t^{2}} + \frac{y^{2} + gt^{2}y + \frac{1}{4}g^{2}t^{4}}{t^{2}}$
$V^{2} = \frac{x^{2}}{t^{2}} + \frac{y^{2} + gt^{2}y + \frac{1}{4}g^{2}t^{4}}{t^{2}}$
$t^{2}V^{2} = x^{2} + y^{2} + gt^{2}y + \frac{1}{4}g^{2}t^{4}$
$4t^{2}V^{2} = 4x^{2} + 4y^{2} + 4gt^{2}y + g^{2}t^{4}$
$4x^{2} + 4y^{2} + 4gt^{2}y + g^{2}t^{4} - 4t^{2}V^{2} = 0$
$4y^{2} + 4gt^{2}y + (gt^{2}t^{4} + 4x^{2} - 4t^{2}V^{2}) = 0$

(ii): Impact is when x = d.
$\text{Impact: } d = Vtcos\theta$
$t = \frac{d}{Vcos\theta}$

First impact occurs when t is minimised.
This means that Vcos theta is maximised, which means cos theta = 1, and theta = 0
$\therefore \text{First impact occurs at } \theta = 0\text{: }t = \frac{d}{V(1)} = \frac{d}{V}$

i'll do the rest later.

8 0

3 years ago