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alisha [4.7K]
3 years ago
5

C+4<-1 Tell whether -6 is solution of the inequality

Mathematics
1 answer:
Lorico [155]3 years ago
5 0

Answer: \begin{bmatrix}\mathrm{Solution:}\:&\:C

Step-by-step explanation:

C+4

C+4-4

C

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Solve and explain your reasoning (x+3)=-6​
densk [106]

Answer:

(x+3)=-6

x=-6-3

x=-9

Step-by-step explanation:

Hope it helps!

When side changes positive changes into negative.

5 0
3 years ago
Find the measures of the angles of a triangle if the measure of one angle is twice the measure of a second angle and the third a
guapka [62]

Answer:

<h2><em> 38°, 66° and 76°</em></h2>

Step-by-step explanation:

A triangle consists of 3 angles and sides. The sum of the angles in a triangle is 180°. Let the angle be <A, <B and <C.

<A + <B + <C = 180° ...... 1

If the measure of one angle is twice the measure of a second angle then

<A = 2<B ...... 2

Also if the third angle measures 3 times the second angle decreased by 48, this is expressed as <C = 3<B-48............ 3

Substituting equations 2 and 3 into 1 will give;

(2<B) + <B + (3<B-48) = 180°

6<B- 48 = 180°

add 48 to both sides

6<B-48+48 = 180+48

6<B = 228

<B = 228/6

<B =38°

To get the other angles of the triangle;

Since <A = 2<B  from equation 2;

<A = 2(38)

<A = 76°

Also <C = 3<B-48 from equation 3;

<C = 3(38)-48

<C = 114-48

<C = 66°

<em>Hence the measures of the angles of the triangle are 38°, 66° and 76°</em>

4 0
3 years ago
HEEEELP <br> ABC m a-58<br> b-42<br> c-55<br> d-41
Savatey [412]
Your answer is going to be C
7 0
3 years ago
Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 15 grams
hram777 [196]

Answer:

X(16)=25.71grams

Step-by-step explanation:

let X(t) denote grams of C formed in  t mins.

For X grams of C we have:

\frac{2}{3}Xg of A and \frac{1}{3}Xg of B

Amounts of A,B remaining at any given time is expressed as:

40-\frac{2}{3}Xg of A and  50-\frac{1}{3}Xg  of B

Rate at which C is formed satisfies:

\frac{dX}{dt} \infty(40-\frac{2}{3}X)(50-\frac{1}{3}X)->\frac{dX}{dt}=k(90-X)\\\therefore \frac{dX}{(90-X)^2}=kdt->\int{\frac{dX}{(90-X)^2}} \, =\int {k} \, dt  \\\therefore \frac{1}{90-X}=kt+c->90-X=\frac{1}{kt+c}\\\\X(t)=90-\frac{1}{kt+c}

Apply the initial condition,X(0)=0 ,to the expression above

0=90-\frac{1}{c} \ \ ->c=\frac{1}{90}\\\therefore\\X(t)=90-\frac{1}{kt+\frac{1}{90}} \ \ ->X(t)=90-\frac{90}{90kt+c}

Now at X(8)=15:

15=90-\frac{90}{90\times 8k+1}  \ ->75=\frac{90}{720k+1}\\k=0.0002778

Substitute  in X(t) to get

X(t)=90-\frac{90}{0.0002778t\times 90+1}\\X(t)=90-\frac{90}{0.25t+1}\\But \ t=16\\\therefore X(t)=90-\frac{90}{0.025\times16+1}\\X(t)=25.71

5 0
3 years ago
2 poi
jolli1 [7]

Me ayudas por favor es regla de ruffini apretá mi perfil y ve mi pregunta

7 0
3 years ago
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