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BaLLatris [955]
3 years ago
14

Simplify the following expression (30/45g+20)-(8/9g+7)

Mathematics
2 answers:
Nata [24]3 years ago
6 0

117 - 2g

 ————————

    9    

olga_2 [115]3 years ago
4 0

\frac{30}{45}g +20 - \frac{8}{9}g + 7 = \frac{15}{9}g + 27 -\frac{8}{9}g = \frac{7}{9}g+27

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30 POINTS I NEED HELP PLZ
IRISSAK [1]

Answer:

120

Step-by-step explanation:

to find the volume you just do length x width x height.

7 0
3 years ago
The ellipse with center C(2, -3), vertices V(-8,-3) and V2(12,-3), and foci F1(-4,-3) and F2(8,-3). 6.
attashe74 [19]

Answer:

(x-2)²/100 + (y+3)²/64 = 1

Step-by-step explanation:

C (2,-3): h=2 k=-3

semimajor axis (CV): a=12-2=10

center-foci: c=8-2=6

semi minor axis: b² = a²-c²= 100 - 36 = 64

equation: (x-h)²/a² + (y-k)²/b² = 1

(x-2)²/100 + (y+3)²/64 = 1

6 0
3 years ago
Steve kicks a football into the air. The path that a ball takes can be described by the equation h=20t - 5t squared where h is t
FinnZ [79.3K]
Since h represents the height of the ball at any given time, t, let h = 25, such that the ball will be 25m high at t.

Now, we have 25 = 20t - 5t²
5t² - 20t + 25 = 0
t² - 4t + 5 = 0

\frac{4 +- \sqrt{16 - 20}}{2}

Since the discriminant is less than zero, there are no solutions.
Hence, the ball will never be 25m high.

4 0
3 years ago
What is the equation of the line perpendicular to 2x − 5y = −35 that contains the point (10, 4)?
Gnoma [55]
Perpendicular has slope that multiplies to -1 in other sloope

thsi one

2x-5y=-35
-5y=-2x-35
y=2/5x+7
slope=2/5
2/5 times -5/2=-1
y=-5/2x+b
findn b
(10,4)
(x,y)
4=-5/2(10)+b
4=-25+b
21=b
y=-5/2x+21
2y=-5x+42
5x+2y=42
7 0
3 years ago
Read 2 more answers
Find the angle between the given vectors to the nearest tenth of a degree. u = <-5, -4>, v = <-4, -3> (1 point)
sesenic [268]

Angle between u = -5i-4j , v=-4i-3j is x =0° .

<u>Step-by-step explanation:</u>

We have , two vectors u = <-5, -4>, v = <-4, -3>  or , u = -5i-4j , v=-4i-3j

We need to find angle between these two vectors . Let's find out:

We know that dot product of two vectors is defined as :

u.v =|u|(|v|)cosx , where x is angle between u & v !

⇒ u.v =|u|(|v|)cosx

⇒ cosx =\frac{u.v}{|u|(|v|)}

Now , u.v = (-5i-4j)(-4i-3j)

⇒ u.v = (-5i-4j)(-4i)-(-5i-4j)(3j)

⇒ u.v = 20+12            { i(j) = j(i) =0  }

⇒ u.v = 32

Now , Modulus of any vector  r = xi+yj is |r| = \sqrt{x^{2}+y^{2}} So ,

|u| = \sqrt{(-5)^{2}+(-4)^{2}} = \sqrt{25+16} = \sqrt{41} \\\\|v| = \sqrt{(-4)^{2}+(-3)^{2}} = \sqrt{16+9} = \sqrt{25} = 5

Putting all these values in equation cosx =\frac{u.v}{|u|(|v|)} we get:

⇒ cosx =\frac{32}{5(\sqrt{41})}

⇒ cos^{-1}(cosx) =cos^{-1}(\frac{32}{5(6.4)})

⇒ x =cos^{-1}(1)                 { cos0 = 1  }

⇒ x =0°

Therefore , Angle between u = -5i-4j , v=-4i-3j is x =0° .

8 0
3 years ago
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