All categories

3 years ago

Select all the partial products for

Mathematics

2 answers:

babymother [125]3 years ago

8 0

Answer:

ok639010 is the answer of the day when I am in the toilet

m_a_m_a [10]3 years ago

6 0

Answer:

try conections school during 2020 its the beast its virtual school all the time

Step-by-step explanation:

chill

You might be interested in

3x - 7y = -40 <br><br> 3x - 2y = -4

eimsori [14]

Answer:

What are you solving for?

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Explain please and tell me how to do it ?

SCORPION-xisa [38]

The trick to calculating the area here is to subdivide the diagram into smaller parts. For example, there's a 3 cm-by-3cm square. The rectangle is 4 cm by 8 cm.. The triangle has a base of 5 cm and a height of 3 cm.

Total area = area of square + area of rectangle + area of triangle

= 9 cm^2 + 32 cm^2 + (1/2)(5 cm)(3 cm)

= (9 + 32 + 7.5) cm^2

= 48.5 cm^2 (total area of figure)

8 0

3 years ago

Read 2 more answers

I need help<br><br>I need u to answer this question. what's 2+2????

umka21 [38]

Answer:

2+2=4

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

(1-сos mx)

Maru [420]

22xsgjkbcgllllllllll

7 0

3 years ago

The function f(x) = 2x2 + 3x + 5, when evaluated, gives a value of 19. What is the function’s input value?

goldenfox [79]

$f(x)=\quad 2{ x }^{ 2 }+3x+5$

Let's called the input 'z'

When we plug 'z' in the function we get ;

$f(z)=\quad 2{ z }^{ 2 }+3z+5$

And we know that, this is equal to 19, so ;

$2{ z }^{ 2 }+3z+5=\quad 19$

Let's rearrange the equation.

$2{ z }^{ 2 }+3z+5=\quad 19\\ \\ 2{ z }^{ 2 }+3z=\quad 19-5\\ \\ 2{ z }^{ 2 }+3z=\quad 14\\ \\ 2{ z }^{ 2 }+3z-14=\quad 0$

So we have a quadratic equation here.

We'll use this formula to solve it :

$\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

The formula is used in equation formed like this :

$a{ x }^{ 2 }+bx+c=0$

In our equation,

a=2 , b=3 and c=-14

Let's plug in the values in the formula to solve,

$a=2\quad b=3\quad c=-14\\ \\ \frac { -3\pm \sqrt { 9-(4\cdot 2\cdot -14) } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 9-(-112) } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 9+112 } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 121 } }{ 4 } \\ \\ \frac { -3\pm 11 }{ 4 }$

So,

$z=\quad \frac { -3+11 }{ 4 } \quad ,\quad \frac { -3-11 }{ 4 } \\ \\ z=\quad \frac { 8 }{ 4 } \quad ,\quad \frac { -14 }{ 4 } \\ \\ z=\quad 2,\quad -\frac { 7 }{ 2 }$

So the input can be both, 2 and $-\frac{7}{2}$

8 0

3 years ago