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timama [110]
3 years ago
13

Find the roots. 2x^2 + 4x - 5 =0

Mathematics
1 answer:
xz_007 [3.2K]3 years ago
8 0

Answer:

-1\pm\dfrac{\sqrt{14}}{2}

Step-by-step explanation:

2x^2+4x-5=0

Using the quadratic formula:

x=\dfrac{-4\pm \sqrt{16+40}}{4}=-1\pm\dfrac{\sqrt{14}}{2}

Hope this helps!

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7.Find the lengths of the missing sides in the triangle. If your answer is not an integer, leave it in simplest radical form. Th
Fed [463]

Here a right angled triangle given. one angle with measure 45^o given. The three sides of the triangle given 4, x, y.

We have to find, the sides which is opposite, adjacent and hypotenuse here.

We know that the side opposite to right angle is always hypotenuse. So, hypotenuyse = y.

The side adjacent to the given angle 45^o is x. So, here adjacent = x.

The opposite side is opposite to the given angle. So, opposite = 4.

Now we will use SOHCAHTOA that is sin(x) =\frac{Opposite}{Hypotenuse} , cos(x) =\frac{Adjacent}{Hypotenuse} , tan(x) =\frac{Opposite}{Adjacent}, where x is the angle given.

To get x, we will use tan. So we will get,

tan(45^o) = \frac{4}{x}

We know the value of tan(45^o) = 1. By substituting the value we will get,

1 =\frac{4}{x}

To find x, we have to move x here to the left side by multiplying it to both sides. We will get,

(1)(x) = (\frac{4}{x}) (x)

x = 4

So we have got the value of x here.

Now to find y, we will use the trigonometric function sine.

sin(45^o) =\frac{4}{y}

we know the value of sin(45^o) =\frac{\sqrt{2}}{2}

By substituting the value we will get,

\frac{\sqrt{2}}{2}  = \frac{4}{y}

By cross multiplying we will get,

(\sqrt{2}) (y) = (4)(2)

\sqrt{2}y = 8

We will get y by dividing both sides by \sqrt{2}, we will get,

\frac{\sqrt{2}y}{\sqrt{2}}   =\frac{8}{\sqrt{2} }

y =\frac{8}{\sqrt{2}  }

Now we will rationalize the denominator by multiplying \sqrt{2} to the top and bottom.

y =\frac{8\sqrt{2}}{(\sqrt{2})(\sqrt{2})}

y =\frac{8\sqrt{2}}{2}

y = 4\sqrt{2}

So we have got the required values of x and y.

8 0
3 years ago
Read 2 more answers
About how much difference in average leaf width would you find in two forests whose average annual rainfalls are near 1500 mm bu
goblinko [34]

Answer:

4.80 mm to 2 d.p

Step-by-step explanation:

w = f(r)

f'(r) = (dw/dr) = (0.0218 mm/mm)

So, for a small difference in rainfall of 220 mm, what is the corresponding small difference in width of leaves in the two forests given.

One definition of a derivative or a rate of change is that it is the ratio of very small differences in the dependent variable to very small differences in the independent variable.

Mathematically,

(dw/dr) = (Δw/Δr) for very small Δw and Δr.

0.0218 = (Δw/220)

Δw = 0.0218 × 220 = 4.796 mm = 4.80 mm to 2 d.p

Hope this Helps!!!

5 0
3 years ago
A cyclist rides her bike at a speed of 30 kilometers per hour. What is this speed in kilometers per minute? How many kilometers
Nata [24]

Step-by-step explanation:

The answer is mentioned above.

3 0
3 years ago
If the dimensions of a pentagonal prism are quadrupled, then the surface area of the prism is multiplied by eight.
Lubov Fominskaja [6]

Answer:

false

Step-by-step explanation:

the relationship between lengths/dimensions and areas is that areas are created by multiplying 2 dimensions.

when you quadruple (×4) the dimensions, then the areas are growing with the square of the factor (×4×4 = ×16), because the factor goes twice into the multiplication : one time for every dimension involved.

so, quadrupling the dimensions would multiply the areas by 16.

8 0
3 years ago
Help me please ;-;!!!!! Use a coordinate grid to create a map of a town with at least five different locations, such as a house,
Sholpan [36]
A. Our diagram have six location: my house (2,1), a movie theater (5,3), the school (6,6), the police station (3,-2), the airport (-3,3), and the mall (-2,-3)

B. Distance between my house and the movie theater:
To calculate the distance we are going to use the points (2,1) and (5,3) to create a right triangle with tow legs of measures 3 and 2; the hypotenuse of the right triangle will be the distance between my house and the movie theater.
Using the Pythagorean theorem:
d^2=3^2+2^2
d^2=9+4
d^2=13
d= \sqrt{13}
We can conclude that the distance between my house and the movie theater is \sqrt{13}

Our second distance is the distance between the police station and the airport:
This time we are using the points (3,-2) and (-3,3) to create a right triangle with legs of measures 6 and 5; the hypotenuse of our triangle will be the the distance between our points:
d^2=6^2+5^2
d^2=36+25
d^2=61
d= \sqrt{61}
We can conclude that the distance between the police station and the airport is \sqrt{61}

C. Distance form the park to the Fire Station:
d^2=3^2+1^2
d^2=9+1
d^2=10
d= \sqrt{10}
We can conclude that the distance from the park to the Fire Station is \sqrt{10}

Distance from the stadium to the mall:
d^2=2^2+1^2
d^2=4+1
d^2=5
d= \sqrt{5}
We can conclude that the distance from the stadium to the mall is \sqrt{5}

D. What is the distance between your house and the mall?
Answer:
d^2=4^2+4^2
d^2=16+16
d^2=32
d= \sqrt{32}
d=4 \sqrt{2}
We can conclude that the distance between my house and the mall is 4 \sqrt{2}

What is the distance between the movie theater and the school?
answer:
d^2=3^2+1^2
d^2=9+1
d^2=10
d= \sqrt{10}
We can conclude that the distance between the movie theater and the school is \sqrt{10}

7 0
3 years ago
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