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3 years ago

Helpppppppppppppppppppppppppppppppppppppp

Mathematics

1 answer:

cupoosta [38]3 years ago

7 0

Answer: A

Step-by-step explanation: the answer couldn’t be B since the point the graph starts is -2.

The slope of the line is -1/3 because that is how many points you go past to find the next point on the graph. So all of the 3x answers would not work

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(3cosx-4sinx) + (3sinx+4cosx) = 5

Ksenya-84 [330]

(3 cos x-4 sin x)+(3sin x+4 cos x)=5
(3cos x+4cos x)+(-4sin x+3 sin x)=5
7 cos x-sin x=5
7cos x=5+sin x
(7 cos x)²=(5+sinx)²
49 cos²x=25+10 sinx+sin²x
49(1-sin²x)=25+10 sinx+sin²x
49-49sin²x=25+10sinx+sin²x
50 sin² x+10sinx-24=0

Sin x=[-10⁺₋√(100+4800)]/100=(-10⁺₋70)/100
We have two possible solutions:
sinx =(-10-70)/100=-0.8
x=sin⁻¹ (-0.8)=-53.13º (360º-53.13º=306.87)

sinx=(-10+70)/100=0.6
x=sin⁻¹ 0.6=36.87º

The solutions when 0≤x≤360º are: 36.87º and 306.87º.

3 0

3 years ago

Base: 11 m, area: 1115.5 m2

Marina86 [1]

101.4

$1115.5 \div 11 = 101.4$

8 0

3 years ago

Convert 2000 ft./min. to miles per hour round to the nearest 10th

Serggg [28]

Answer:

The conversion of 2000 ft per min into miles per hour is 2.268 miles per hour .

Step-by-step explanation:

Given as :

The speed = s = 2000 ft per min

Now, ∵ 5280 feet = 1 miles

so, 1 feet = $\dfrac{1}{5280}$ miles

∴ 2000 feet = $\dfrac{1}{5280}$ × 2000 miles

i.e 2000 feet = 0.378 miles

<u>Again</u>

∵ 1 minute = $\dfrac{1}{60}$ hour

So, The speed in miles per hour

s = $\dfrac{distance}{time}$

∴ s = $\frac{0.378}{\frac{1}{60}}$

i.e s = 2.268 miles per hour

Hence, The conversion of 2000 ft per min into miles per hour is 2.268 miles per hour . answer

8 0

3 years ago

Find the distance of a line given the following points. R(3, 1), W(-1, -2)

olga_2 [115]

Answer:

5 units

Step-by-step explanation:

±√(- 2 - 1)^2 + (- 1 - 3)^2 = ±5 (rej - 5 since distance > 0)

Apply distance formula

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7 0

3 years ago

Which function represents a translation of the parent cubic function 8 units to the left? h(x) = x3 + 8 h(x) = x3 – 8 h(x) = (x

Ksju [112]

Using the definition of the Vertical shifts of graphs of the function :

"Suppose c>0,

To graph y=f(x)+c, shift the graph of y=f(x) upward c units.

To graph y=f(x)-c, shift the graph of y=f(x) downward c units"

Again we recall the definition of Horizontal shifts of graphs:

" suppose c>0,

the graph y=f(x-c), shift the graph of y=f(x) to the right by c units

the graph y=f(x+c), shift the graph of y=f(x) to the left by c units. "