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4 years ago

14

SOLVE 3x²-7x-12=0 correct answer to 2 decimal places

1 answer:

Bas_tet [7]4 years ago

5 0

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I do not understand what I am supposed to do here

notka56 [123]

You plug in 25 for "P", and 7 for "a" and "b", then you find "c". The answer is 11.

7 0

3 years ago

Does anyone know what is the value of x

ANTONII [103]

Answer:

7.5

Step-by-step explanation:

By intersecting secants theorem:

$12(12 - x) = 6(6 + 3) \\ \\ 12(12 - x) = 54 \\ \\ 144 - 12x = 54 \\ \\ 144 - 54 = 12x \\ \\ 90 = 12x \\ \\ x = \frac{90}{12} \\ \\ x = 7.5$

4 0

3 years ago

Find the diagonal of the rectangular solid with the given measures part of the answer is provided for you

Svet_ta [14]

Answer: $\sqrt{59}$

Step-by-step explanation:

The diagonal of a rectangular prism can be found with $\sqrt{l^2+w^2+h^2}$

$l^2+w^2+h^2=59$ . Thus, the diagonal is $\sqrt{59}$

Hope it helps <3

3 0

4 years ago

What is the length of the segment, endpoints of which are intersections of parabolas y=x^2− 11/4 x− 7/4 and y=− 7/8 x^2+x+ 31/8

kobusy [5.1K]

Answer:

The length of line segment is 5

Step-by-step explanation:

we are given equation of parabolas as

$y=x^2-\frac{11}{4}x-\frac{7}{4}$

$y=-\frac{7}{8}x^2+x+\frac{31}{8}$

Firstly, we will find intersection points

we can set them equal

and then we can solve for x

$x^2-\frac{11}{4}x-\frac{7}{4}=-\frac{7}{8}x^2+x+\frac{31}{8}$

Multiply all sides by 8

$x^2\cdot \:8-\frac{11}{4}x\cdot \:8-\frac{7}{4}\cdot \:8=-\frac{7}{8}x^2\cdot \:8+x\cdot \:8+\frac{31}{8}\cdot \:8$

$8x^2-22x-14=-7x^2+8x+31$

$15x^2-30x-45=0$

now, we can factor it

$15(x^2-2x-3)=0$

$15(x-3)(x+1)=0$

$x=-1,x=3$

now, we can find y-values

At x=-1:

$y=(-1)^2-\frac{11}{4}(-1)-\frac{7}{4}$

y=2

so, we get point as

(-1,2)

At x=3:

$y=(3)^2-\frac{11}{4}(3)-\frac{7}{4}$

y=-1

so, we get point as

(3,-1)

now, we can find distance between these two points

(-1,2)

x1=-1 , y1=2

(3,-1)

x2=3 , y2=-1

now, we can find distance

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

now, we can plug values

$D=\sqrt{(3+1)^2+(-1-2)^2}$

$D=5$

So,

The length of line segment is 5

6 0

3 years ago

Am I correct on question #6?

pashok25 [27]

Option A:

$\tan(105^\circ)=-(2+\sqrt{3})$

Solution:

<u>To evaluate tan(105)°:</u>

105° can be written as sum of 60° and 45°.

tan(105)° = tan(45 + 60)°

Using the summation identity:

$$\tan (x+y)=\frac{\tan (x)+\tan (y)}{1-\tan (x) \tan (y)}$

$$\tan \left(105^{\circ}\right)=\frac{\tan \left(45^{\circ}\right)+\tan \left(60^{\circ}\right)}{1-\tan \left(45^{\circ}\right) \tan \left(60^{\circ}\right)}$

We know that, tan(45)° = 1 and tan(60)° = √3

Substitute this in the above equation.

$$=\frac{1+\sqrt{3}}{1-1 \cdot \sqrt{3}}$

$$=\frac{1+\sqrt{3}}{1-\sqrt{3}}$

To rationalize the denominator multiply by the conjugate $\frac{1+\sqrt{3}}{1+\sqrt{3}}$ .

$$=\frac{(1+\sqrt{3})(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}$

Using exponent formula: $a^{b} \cdot a^{c}=a^{b+c}$ and $(x-y)(x+y)=x^2-y^2$

$$=\frac{(1+\sqrt{3})^2}{(1^2-(\sqrt{3})^2)}$

Using exponent formula: $(a+b)^{2}=a^{2}+2 a b+b^{2}$

$$=\frac{1^{2}+2 \cdot 1 \cdot \sqrt{3}+(\sqrt{3})^{2}}{1-3}$

$$=\frac{4+2 \sqrt{3}}{-2}$

$$=\frac{2(2+ \sqrt{3})}{-2}$

$=-(2+\sqrt{3})$

$\tan(105^\circ)=-(2+\sqrt{3})$

Hence option A is the correct answer.

6 0

4 years ago