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belka [17]
3 years ago
13

I GIVE BRAINLIEST MARKS I SWEAR rational functions, construct table values and sketch the graph of rational functions

Mathematics
2 answers:
Lostsunrise [7]3 years ago
5 0

Answer:Hope this is rght I only can answer the first

one.

Step-by-step explanation:

Marianna [84]3 years ago
3 0

Answer:

Download math)way it helps I swear

Step-by-step explanation:

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X : 14= 7: 8<br> can someone solve this for me and explain me
Vanyuwa [196]

x:14 = 7:8

x:14 = x/14

7:8 = 7/8

x/14 = 7/8

Multiply the numerator of the 1st ratio with the denominator of the 2nd.

And multiply the numerator of the 2nd ratio with the denominator of the 1st.

Doing so, we get -

x * 8 = 7 * 14

So, 8x = 7 * 14

8x = 98

We divide 98 by 8 (transposition method)

x = 98/8

Therefore, x = 12.25

6 0
3 years ago
Eight people were asked, "How many times have you been on an airplane?" Their responses are listed below.
laila [671]
Mode: 1

Range: 5
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4 0
3 years ago
12345 :- 987 me pueden ayudar con esta division
Sophie [7]
12.50 espero que te sirva :)
4 0
3 years ago
List the elements in the set.<br> 1) {x 1 x is an integer between -7 and -3}
omeli [17]

Answer:

सामाजिक अध्ययन र जिवनोपयोगी शिक्षाको महत्व दसांउनुहोस

8 0
1 year ago
A particle is projected with a velocity of <img src="https://tex.z-dn.net/?f=40ms%5E-%5E1" id="TexFormula1" title="40ms^-^1" alt
Katena32 [7]

Answer:

2\sqrt{55}\text{ m/s or }\approx 14.8\text{m/s}

Step-by-step explanation:

The vertical component of the initial launch can be found using basic trigonometry. In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. Let the vertical component at launch be y. The magnitude of 40\text{ m/s} will be the hypotenuse, and the relevant angle is the angle to the horizontal at launch. Since we're given that the angle of elevation is 60^{\circ}, we have:

\sin 60^{\circ}=\frac{y}{40},\\y=40\sin 60^{\circ},\\y=20\sqrt{3}(Recall that \sin 60^{\circ}=\frac{\sqrt{3}}{2})

Now that we've found the vertical component of the velocity and launch, we can use kinematics equation v_f^2=v_i^2+2a\Delta y to solve this problem, where v_f/v_i is final and initial velocity, respectively, a is acceleration, and \Delta y is distance travelled. The only acceleration is acceleration due to gravity, which is approximately 9.8\:\mathrm{m/s^2}. However, since the projectile is moving up and gravity is pulling down, acceleration should be negative, represent the direction of the acceleration.

What we know:

  • v_i=20\sqrt{3}\text{ m/s}
  • a=-9.8\:\mathrm{m/s^2}
  • \Delta y =50\text{ m}

Solving for v_f:

v_f^2=(20\sqrt{3})^2+2(-9.8)(50),\\v_f^2=1200-980,\\v_f^2=220,\\v_f=\sqrt{220}=\boxed{2\sqrt{55}\text{ m/s}}

3 0
3 years ago
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