All categories

3 years ago

Can someone please help me

Mathematics

2 answers:

pentagon [3]3 years ago

7 0

The first one is 8 and here's a graph. the second one is 5 again here's a graph

Oxana [17]3 years ago

4 0

2x ≤ 16
Divide by 2.
x ≤ 8

-3x + 7 < -8
Subtract 7.
-3x < -15
Divide by -3 and flip the inequality symbol.
x > 5

You might be interested in

Find the value of x. Show your work.

yarga [219]

Answer:

x = 11

Step-by-step explanation:

∠BCE + ∠ECF = 180°

∴(10x + 15)° + 5x° = 180°

(15x + 15)° = 180°

15x = 180° - 15°

15x = 165°

x = 165°/15°

x = 11°

4 0

4 years ago

Please solve this question?

Alchen [17]

Given,

$\lim_{x}^{a}\dfrac{x^{2/3}-a^{2/3}}{x-a}$

We can use L'Hopital's Rule to get,

$\lim_{x}^{a}\dfrac{2}{3-\sqrt[3]{x}}$

Now plug in a,

$\boxed{\dfrac{2}{3-\sqrt[3]{a}}}$

Hope this helps.

r3t40

3 0

4 years ago

6+2i

aliya0001 [1]

$\frac{6+2i}{3+3i}=\frac{6+2i}{3+3i}\times\frac{3-3i}{3-3i}$
$\frac{6+2i}{3+3i}=\frac{18-18i+6i-6i^{2}}{9-9i+9i-9i^{2}}$
$\frac{6+2i}{3+3i}=\frac{18-12i+6}{9+9}$
$\frac{6+2i}{3+3i}=\frac{24-12i}{18}$
$\frac{6+2i}{3+3i}=\frac{12(2-i)}{18}$
$\frac{6+2i}{3+3i}=\frac{2(2-i)}{3}$
$\frac{6+2i}{3+3i}=\frac{2}{3}(2-i)$
$\frac{6+2i}{3+3i}=\frac{4}{3}-\frac{2i}{3}$

3 0

4 years ago

In rectangle QRST, diagonals QS and RT are drawn and intersect at point P. Which of the following st

natta225 [31]

Answer:

Step-by-step explanation:

8 0

3 years ago

Select the best answer for the question (Algebra II)

Rainbow [258]

The answer is A, simple math with a calculator

4 0

4 years ago