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bekas [8.4K]
3 years ago
15

Evaluate 1 3/7-(2/3+5/7)​

Mathematics
2 answers:
Artyom0805 [142]3 years ago
7 0

Answer:

13/7 - (14/21 + 15/21)

Step-by-step explanation:

= 13/7 - ( 29/21)

= 39/21 - 29/21 = 10/21 = 0.47

KATRIN_1 [288]3 years ago
6 0

Answer:

that's the answer .......

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11 cm<br> 7 cm<br> 4 cm<br> What is the volume of the right rectangular prism?
netineya [11]
Hi, just do 11x7x4cm.
5 0
3 years ago
P is a rectangle with a length 40 cm and width x cm
romanna [79]
We know that

the length of Q is 25% more than the length of p
so
length Q=1.25*40--------> 50 cm
the area of Q is 10% less than the area of p
Area Q=50*y
Area P=40*x
so
50*y=0.90*[40*x]---------> 50*y=36*x-------> x/y=50/36---> 25/18

the answer 
the ratio x:y is 25:18
4 0
3 years ago
Find the perimeter of the following trapezoid:<br>6ft<br>8ft<br>2.5ft<br>2.5ft<br>2ft​
kirill [66]

Answer:

21

Step-by-step explanation:

Add all the numbers together

6+8+2.5+2.5+2=21

6 0
3 years ago
Suppose a football is kicked with an initial velocity of 82 ft/sec., at an angle of
satela [25.4K]

Answer:

The position P is:

P = 87\^x + 75\^y ft     <u><em> Remember that the position is a vector. Observe the attached image</em></u>

Step-by-step explanation:

The equation that describes the height as a function of time of an object that moves in a parabolic trajectory with an initial velocity s_0 is:

y(t) = y_0 + s_0t -16t ^ 2

Where y_0 is the initial height = 0 for this case

We know that the initial velocity is:

82 ft/sec at an angle of 58 ° with respect to the ground.

So:

s_0 = 82sin(58\°) ft/sec

s_0 = 69.54 ft/sec

Thus

y(t) = 69.54t -16t ^ 2

The height after 2 sec is:

y(2) = 69.54 (2) -16 (2) ^ 2

y(2) = 75\ ft

Then the equation that describes the horizontal position of the ball is

X(t) = X_0 + s_0t

Where

X_ 0 = 0 for this case

s_0 = 82cos(58\°) ft / sec

s_0 = 43.45 ft/sec

So

X(t) = 43.45t

After 2 seconds the horizontal distance reached by the ball is:

X (2) = 43.45(2)\\\\X (2) = 87\ ft

Finally the vector position P is:

P = 87\^x + 75\^y ft

6 0
3 years ago
write a polynomial function of least degree with integral coefficients that has the given zeros. -(1/3), -i
inessss [21]

Answer:

f(x)=3x^3+x^2+3x+1

Step-by-step explanation:

If a real number -\frac{1}{3} is a zero of polynomial function, then

x-\left(-\dfrac{1}{3}\right)=x+\dfrac{1}{3}

is the factor of this function.

If a complex number -i is a xero of the polynomial function, then the complex number i is also a zero of this function and

x-(-i)=x+i\ \text{ and }\ x-i

are two factors of this function.

So, the function of least degree is

f(x)=\left(x+\dfrac{1}{3}\right)(x+i)(x-i)=\left(x+\dfrac{1}{3}\right)(x^2-i^2)=\\ \\ =\left(x+\dfrac{1}{3}\right)(x^2+1)=\dfrac{1}{3}(3x+1)(x^2+1)=\dfrac{1}{3}(3x^3+x^2+3x+1)

If the polynomial function must be with integer coefficients, then it has a form

f(x)=3x^3+x^2+3x+1

4 0
3 years ago
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