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3 years ago

What is the answer to 19/5+5/4

Mathematics

2 answers:

elena-s [515]3 years ago

7 0

Answer:

exact form : 101/20

Decimal : 5.05

Mixed Number: 5 1/20

Furkat [3]3 years ago

4 0

The answer would be 101/20

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work out the circumference of this circle take (pie) to be 3.142 and give ur answer in 1 decimal place radius=9m

Whitepunk [10]

Answer:

56.6m to 1.d.p

Step-by-step explanation:

circumference=2πr

=2×3.142×9

=56.556m

=56.6m to 1.d.p

4 0

3 years ago

Which type of transformation is described by (x, y) -> + (x + 3, y — 2)?

creativ13 [48]

Answer:

(x, y) → (x+3, y-2) means Translation 3 units left and 2 units down.

Step-by-step explanation:

Given the rule of transformation:

(x, y) → (x+3, y-2)

If we move horizontally 3 units to the right, 3 is added to the x-coordinate of any given point, and then 2 units down.

In other words,

(x, y) → (x+3, y-2) means Translation 3 units left and 2 units down.

For example,

Let say if we have the point A(4, 5)

Using the translation rule:

(x, y) → (x+3, y-2)

A(4, 5) → A'(4 + 3, 5 - 2) → A'(7, 3)

Therefore, the new position of A(4, 5) after translating 3 units left and 2 units down would be: A'(7, 3)

4 0

3 years ago

How to change .365 to a percent

KATRIN_1 [288]

Multipl .365 by 100 and you get 36.5%

3 0

3 years ago

Read 2 more answers

These Are the answer choices: A.) 343 B.) 342 C.) 432 D.) 677

Tanya [424]

The answer is C.) 432

4 0

3 years ago

Determine the values of the constants b and c so that the function given below is differentiable. f(x)={2xbx2+cxx≤1x>1

Lera25 [3.4K]

Assuming the function is

$f(x)=\begin{cases}2x&\text{for }x\le1\\bx^2+cx&\text{for }x>1\end{cases}$

For $f(x)$ to be differentiable, it necessarily has to be continuous. For this condition to be met, we need

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)$
$\iff\displaystyle\lim_{x\to1}2x=\lim_{x\to1}(bx^2+cx)$
$\iff2=b+c$

For the derivative to exist, the one-sided limits of the derivative must also exist and be equal. We have

$f'(x)=\begin{cases}2&\text{for }x1\end{cases}$

$\displaystyle\lim_{x\to1^-}2=\lim_{x\to1^+}(2bx+c)$
$\iff2=2b+c$

Now we solve for $b$ and $c$ :

$\begin{cases}b+c=2\\2b+c=2\end{cases}\implies b=0,c=2$

5 0

4 years ago