All categories

2 years ago

2. If AD= 2x – 1 and BC = 5x – 13, then AD= BC= AB= DC =

Mathematics

1 answer:

Sladkaya [172]2 years ago

4 0

Answer:

the question is missing a relation between AD AND BC

You might be interested in

Find f(x)+g(x), given f(x)=x^2-8 and g(x)=-2-x^2

Gnoma [55]

$\left(x^2-8\right)+\left(-2-x^2\right)=-10$

$\left(x^2-8\right)+\left(-2-x^2\right)$

$=x^2-8-2-x^2$

$=x^2-x^2-8-2$

$=-8-2$

$=-10$

8 0

2 years ago

Jake map shows the distance from the big star coffee to the restaurant supply store as it 3 cm is the scale of the map is 1 cm t

kumpel [21]

Answer: The real distance from the shop to the store = 24 km.

Step-by-step explanation:

Given: The distance from the big star coffee to the restaurant supply store = 3cm (On map)

Scale of the map : 1 cm to 8 km

That means, the real distance from the shop to the store = 3 x ( 8 km ) = 24 km

Hence, the real distance from the shop to the store = 24 km.

7 0

2 years ago

NEED HELP ASAP PLEASE

ki77a [65]

D.) is the right answer

5 0

3 years ago

Read 2 more answers

If f(x)=2x^2+(1000/x), find the average rate of change of f(x) from x=a to x=a+h.

galina1969 [7]

Answer:

$\frac{\frac{1000}{x+h}-\frac{1000}{x}}{h}$ is your average rate of change,

Step-by-step explanation:

average rate of change is

$\frac{f(a+h)-f(a)}{a+h-a}$ , by slope formula

simplify this to get $\frac{f(a+h)-f(a)}{h}$ , which is the definition of the derivative as h goes to 0

$\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$

since you defined x=a, we can substitute a for x and vice versa to find our derivative.

$\lim_{h \to 0} \frac{(2x^2+\frac{1000}{x+h})-(2x^2+\frac{1000}{x})}{h}$

simplifying

$\lim_{h \to 0} \frac{\frac{1000}{x+h}-\frac{1000}{x}}{h}$ (your average rate of change)

6 0

2 years ago

Help find the domain and the range

erica [24]

Answer:

Step-by-step explanation:

domain:-all real numbers.

as y≥-2

so range is all real values ≥-2

6 0

3 years ago