Answer:
3 and 12
Step-by-step explanation:
3*5=15 and 15-3=12. The product of 3 and 12 is 36.
The integers are 3 and 12.
What is Algebra?Algebra is a branch of mathematics dealing with symbols and the rules for manipulating those symbols.
let the one integer be x.
Another integer = 5x - 3
According to question,
x *(5x-3 ) =36
5x ² -3 x =36
5x² -3 x - 36 = 0
5x² -15 x +12 x - 36 =0
5x( x-3) + 12 ( x -3) =0
(5x +12)( x- 3)=0
x= -12/5 or 3
So, the integers are 3 and 12.
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What is f(4) for the function f(x)= 6x+7?
Answer:
31
Step-by-step explanation:
f(4) = 6x4 + 7
= 24 + 7
= 31
A communications tower is supported by two wires, connected at the same point on the ground. One is attached to the tower at D and the long-on at C. The angle AD makes with the ground is 30 ° and the angle between the two wires is 10 °. How much below the top of the tower is the shorter one attached?
Answer:
10.99 m
Step-by-step explanation:
✍️What we are basically asked to solve here is to find the distance between C and D.
To find CD, find the length of BC, and BD. Their difference will give us CD.
Thus, BC - BD = CD.
✍️Finding BC using trigonometric ratio formula:
[tex] \theta = 30 + 10 = 40 [/tex]
Opposite side = BC = ??
Adjacent side = 42 m
Thus:
[tex] tan(\theta) = \frac{opposite}{adjacent} [/tex]
Plug in the values
[tex] tan(40) = \frac{BC}{42} [/tex]
Multiply both sides by 42
[tex] tan(40) \times 42 = \frac{BC}{42} \times 42 [/tex]
[tex] 35.24 = BC [/tex]
BC = 35.24 m
✍️Finding BD using trigonometric ratio formula:
[tex] \theta = 30 [/tex]
Opposite side = BD = ??
Adjacent side = 42 m
Thus:
[tex] tan(\theta) = \frac{opposite}{adjacent} [/tex]
Plug in the values
[tex] tan(30) = \frac{BD}{42} [/tex]
Multiply both sides by 42
[tex] tan(30) \times 42 = \frac{BD}{42} \times 42 [/tex]
[tex] 24.25 = BD [/tex]
BD = 24.25 m
✍️How much below the top of the tower is the shorter one attached:
Thus,
BC - BD = CD
35.24 m - 24.25 m = 10.99 m
A wave is traveling with a speed of 500 m/sec and has a wavelength of 10 m. Determine its frequency.
Answer:
I'm gonna get more people here
Answer:
50 hz yw pplz
Step-by-step explanation:
A car and a truck leave the same intersection, the truck heading north at 60 mph and the car heading west at 55 mph. At what rate is the distance between the car and the truck changing when the car and the truck are 30 miles and 40 miles from the intersection, respectively?
Answer: -81
Step-by-step explanation:
Let x = the car moving towards west = 30
Let y = truck moving towards north = 40
dx/dt = -55
dy/dt = -60
We the use Pythagoras theorem to solve further where,
s² = x² + y²
s² = 30² + 40²
s² = 900 + 1600
s² = 2500
s =✓2500
s = 50
We then differentiate with respect to t, then we will have:
s² = x² + y²
2s ds/dt = 2x ds/dt + 2y dy/dt
50 ds/dt = (30 × -55) + (40 × -60)
50 ds/dt = -1650 - 2400
50 ds/dt = - 4050
ds/dt = -4050/50
ds/dt = -81