All categories

3 years ago

Find the supplement of an angle that measures 89°. A. 61° B. 1° C. 31° D. 91°

Mathematics

1 answer:

Veronika [31]3 years ago

6 0

Answer:

Step-by-step explanation:

Supplementary angles sum to 180°

Subtract the given angle from 180 for the supplement, that is

180° - 89° = 91° ← the supplementary angle

You might be interested in

Negative exponent in a scientific notation means your answer will be negative. <br> True or False

ahrayia [7]

Answer:

false

Step-by-step explanation: negative times a negative is a positive

8 0

4 years ago

You get a 27% discount on a sweater that is $18 sales tax is 8.4% how much is it after text total cost

Oduvanchick [21]

Answer:

27/100×18=4.86

8.4/100×18=1.512

total is 6.372

4 0

3 years ago

If a,b,c and d are positive real numbers such that logab=8/9, logbc=-3/4, logcd=2, find the value of logd(abc)

Eva8 [605]

We can expand the logarithm of a product as a sum of logarithms:

$\log_dabc=\log_da+\log_db+\log_dc$

Then using the change of base formula, we can derive the relationship

$\log_xy=\dfrac{\ln y}{\ln x}=\dfrac1{\frac{\ln x}{\ln y}}=\dfrac1{\log_yx}$

This immediately tells us that

$\log_dc=\dfrac1{\log_cd}=\dfrac12$

Notice that none of $a,b,c,d$ can be equal to 1. This is because

$\log_1x=y\implies1^{\log_1x}=1^y\implies x=1$

for any choice of $y$ . This means we can safely do the following without worrying about division by 0.

$\log_db=\dfrac{\ln b}{\ln d}=\dfrac{\frac{\ln b}{\ln c}}{\frac{\ln d}{\ln c}}=\dfrac{\log_cb}{\log_cd}=\dfrac1{\log_bc\log_cd}$

so that

$\log_db=\dfrac1{-\frac34\cdot2}=-\dfrac23$

Similarly,

$\log_da=\dfrac{\ln a}{\ln d}=\dfrac{\frac{\ln a}{\ln b}}{\frac{\ln d}{\ln b}}=\dfrac{\log_ba}{\log_bd}=\dfrac{\log_db}{\log_ab}$

so that

$\log_da=\dfrac{-\frac23}{\frac89}=-\dfrac34$

So we end up with

$\log_dabc=-\dfrac34-\dfrac23+\dfrac12=-\dfrac{11}{12}$

###

Another way to do this:

$\log_ab=\dfrac89\implies a^{8/9}=b\implies a=b^{9/8}$

$\log_bc=-\dfrac34\implies b^{-3/4}=c\implies b=c^{-4/3}$

$\log_cd=2\implies c^2=d\implies\log_dc^2=1\implies\log_dc=\dfrac12$

Then

$abc=(c^{-4/3})^{9/8}c^{-4/3}c=c^{-11/6}$

So we have

$\log_dabc=\log_dc^{-11/6}=-\dfrac{11}6\log_dc=-\dfrac{11}6\cdot\dfrac12=-\dfrac{11}{12}$

4 0

3 years ago

I need help ASAP!! 13 points

OleMash [197]

Answer:

The correct answer is H

Step-by-step explanation:

1/4 of a wall in 2/3 of an hour is 40 min

so a wall takes 160 min which is 2 hours and 40 min

so 3 walls takes 8 hours

6 0

3 years ago

Read 2 more answers

given y>0 and dy/dx=(3x^2+4x)/yif the point (1,√10) is on the graph relating x and y, then what is y when x=0

jonny [76]

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3x^2+4x}y\iff y\,\mathrm dy=(3x^2+4x)\,\mathrm dx$
$\implies\displaystyle\int y\,\mathrm dy=\int(3x^2+4x)\,\mathrm dx$
$\implies \dfrac12y^2=x^3+2x^2+C$

When $x=1$ you have $y=\sqrt{10}$ , so

$\dfrac12(\sqrt{10})^2=1^3+2(1)^2+C\implies 5=1+2+C\implies C=2$

and so the particular solution to the ODE is

$\dfrac12y^2=x^3+2x^2+2$

Then when $x=0$ , you get

$\dfrac12y^2=0^3+2(0)^2+2=2$
$\implies y^2=4$
$\implies y=2$

where we omitted the negative root because it's given that $y>0$ .

4 0

3 years ago

Find the supplement of an angle that measures 89°. A. 61° B. 1° C. 31° D. 91°​

Find the supplement of an angle that measures 89°. A. 61° B. 1° C. 31° D. 91°