All categories

3 years ago

Which expression is equivalent to -0.8(10.8x-20+3.2x)

1 answer:

6 0

-0.8* 10.8x= -8.64x
-0.8* -20= 16
-0.8* 3.2x =-2.56x

so it is equivalent to the expression -8.64x+16-2.56x but add like terms so it becomes -8.64x-2.56x+16. The final answer is -11.2x=16

Rate and Thanks!

You might be interested in

What is the surface area of a spehere with radius of 13 units

Tju [1.3M]

Answer:

The surface are of a spehere with a radius of 13 is = to 5

Step-by-step explanation:

7 0

3 years ago

HELP!!! WHATS THE ANSWER PLS HELP-

Dmitriy789 [7]

Answer:Adult=105 kid=62

Step-by-step explanation:use x&y

6 0

3 years ago

Please help me ASAP <br> Please help

son4ous [18]

9514 1404 393

Answer:

5 cm

Step-by-step explanation:

The kite is symmetrical about the long axis, so the height of each of the triangles is half the overall width of the kite. (10 cm/2 = 5 cm)

The height of the triangle <em>h</em> is 5 cm.

3 0

3 years ago

The figure below shows three quadrilaterals on a coordinate grid:

Hatshy [7]

Answer:

im just commenting to get points

Step-by-step explanation:

3 0

3 years ago

F(x)=x^2-3x+18 how many distinct real number zeros does f have?

vampirchik [111]

Answer:

None of the distinct real number zeros the function have.

Step-by-step explanation:

Considering the function

$f(x)=x^2-3x+18$

The zeroes of a function are those values that touches the x-axis. In order to find those values we must

Set $f(x) = 0$ in order to find those values

$0=x^2-3x+18$

$\mathrm{Switch\:sides}$

$x^2-3x+18=0$

$\mathrm{Solve\:with\:the\:quadratic\:formula}$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=-3,\:c=18:\quad x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}$

$x=\frac{-\left(-3\right)+\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}$

$x=\frac{3+\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}$

$3+\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}=3+\sqrt{63}i$

$x=\frac{3+\sqrt{63}i}{2\cdot \:1}$

$x=\frac{3+3\sqrt{7}i}{2}$

$x=\frac{3}{2}+\frac{3\sqrt{7}}{2}i$

Similarly,

$x=\frac{-\left(-3\right)-\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}:\quad \frac{3}{2}-i\frac{3\sqrt{7}}{2}$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\frac{3}{2}+i\frac{3\sqrt{7}}{2},\:x=\frac{3}{2}-i\frac{3\sqrt{7}}{2}$

BUT, NONE OF THEM ARE REAL NUMBER ZEROS.

Therefore, none of the distinct real number zeros the function have.

5 0

4 years ago